**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,popcnt,lzcnt,abm,bmi,bmi2,fma,tune=native")
using namespace std;
using namespace __gnu_pbds;

template <typename T>
using o_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using o_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define ll long long
#define ld long double
#define vll vector<ll>
#define vvll vector<vll>
#define vpll vector<pair<ll,ll>>
#define mll map<ll,ll>
#define sll set<ll>
#define pll pair<ll,ll>
#define mod 1000000007
#define mod2 998244353
#define inf 1000000000000000
#define maxN 1e5+1
#define pb push_back
#define loop(i,n) for(ll i=0; i<n; i++)
#define rloop(i,n) for(ll i=n-1; i>=0; i--)
#define loop2(j,start,end) for(ll j=start; j<=end; j++)
#define rloop2(j,start,end) for(ll j=start; j>=end; j--)
#define py cout<<"YES"<<endl
#define pn cout<<"NO"<<endl
#define pnr cout<<"NO"<<endl; return;
#define pyr cout<<"YES"<<endl; return;
#define p(x) cout<<x<<"\n"
#define pl(x) cout<<x<<" "
#define pv(x) loop(i,x.size()) pl(x[i]); cout<<endl;
#define fix(f,n) std::fixed<<std::setprecision(n)<<f

void hi(ll kitna, ll kidharse){
    if(kitna==1){
        cout<<kidharse+1<<" "<<kidharse+1<<endl;
        return;
    }
    hi((kitna-1),kidharse);

    cout<<kidharse<<" "<<kidharse+kitna<<endl;
    cout<<kidharse<<" "<<kidharse+kitna-1<<endl;
    hi((kitna-1),kidharse);
}

void solve(){
    ll n;
    cin>>n;
    ll sum=0;
    vll v(n);
    vll oper={1};
    loop(i,17)  oper.pb(2*oper.back()+2);
    // pv(oper);
    loop(i,n) cin>>v[i];
    loop(i,n) sum+=v[i];
    ll maxi=0,where,op;
    if(n==1){
        if(v[0]!=0) cout<<sum<<" 0"<<endl;
        else{
            cout<<1<<" 1"<<endl;
            cout<<1<<" 1"<<endl;
        }
        return;
    }
    loop(i,(1LL<<n)){
        vll dontchange;
        ll temp=0,tempop=0;
        loop(j,n){
            if((i>>j)&1){
                dontchange.pb(j);
            }else if(v[j]!=0) tempop++;
        }
        ll x=dontchange.size();
        if(x==0){
            temp=n*n;
            if(n>1) tempop+=oper[n-2]+1;
        }
        loop(j,x){
            if(j==0){
                temp+=(dontchange[j]*dontchange[j]);
                if(dontchange[j]>1) tempop+=oper[dontchange[j]-2]+1;
                else if(dontchange[0]==1) tempop++;
            }
            if(j==x-1){
                temp+=(n-dontchange[j]-1)*(n-dontchange[j]-1);
                if((n-dontchange[j]-1)>1) tempop+=oper[(n-dontchange[j]-3)]+1;
                else if((n-dontchange[j]-1)==1) tempop++;
            }
            if(j!=x-1){
                temp+=(dontchange[j+1]-dontchange[j]-1)*(dontchange[j+1]-dontchange[j]-1);
                if((dontchange[j+1]-dontchange[j]-1)>1) tempop+=oper[(dontchange[j+1]-dontchange[j]-3)]+1;
                else if((dontchange[j+1]-dontchange[j]-1)==1) tempop++;
            }
            temp+=v[dontchange[j]];
        }
        if(temp>maxi){
            maxi=temp;
            where=i;
            op=tempop;
        }
    }
    cout<<maxi<<" "<<op<<endl;
    vll dontchange;
    dontchange.clear();
    loop(j,n){
            if((where>>j)&1){
                dontchange.pb(j);
            }else if(v[j]!=0) cout<<j+1<<" "<<j+1<<endl;
        }
    ll x=dontchange.size();
    if(x==0){
        hi(1,1);
        cout<<1<<" "<<n<<endl;
    }else{
        loop(j,x){
            if(j==0){

                if(dontchange[0]>1){
                    hi(dontchange[0]-1,1);
                    cout<<"1 "<<dontchange[0]<<endl;
                }else if(dontchange[0]==1) cout<<"1 1"<<endl;

            }
            if(j==x-1){
                ll num=(n-dontchange[j]-1);
                if(num>1){
                    hi(num-1,dontchange[j]+2);
                    cout<<dontchange[j]+2<<" "<<n<<endl;

                }
                else if((n-dontchange[j]-1)==1) cout<<dontchange[j]+2<<" "<<dontchange[j]+2<<endl;
            }
            if(j!=x-1){
                ll num=(dontchange[j+1]-dontchange[j]-1);
                if(num>1){
                    hi(num-1,dontchange[j]+2);
                    cout<<dontchange[j]+2<<" "<<dontchange[j]+num+1<<endl;
                }else if(num==1){
                    cout<<dontchange[j]+2<<" "<<dontchange[j]+2<<endl;
                }
            }
        }
    }
}

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    ll t=1;
    // cin>>t;
    while(t--) solve();
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that can cause the algorithm to fail. Here are the steps to think through this:

1. **Edge Cases:**
   - Arrays of length 1.
   - Arrays with all elements being the same.
   - Arrays with elements that are already in the optimal state.
   - Arrays with elements that have a large range of values.

2. **Complex Scenarios:**
   - Arrays with multiple segments that can be optimized independently.
   - Arrays where the MEX operation can be applied multiple times in different segments.

3. **Small-Scale Input:**
   - Since the input length is limited to 18, we need to create a test case that fits within this limit but is complex enough to expose potential bugs.

Given these considerations, let's create a test case:

- **Array Length:** 4
- **Array Elements:** A mix of values that can be optimized in different ways.

Here is a small-scale test input:

```plaintext
4
0 1 2 3
```

### Explanation:
- **Initial Array:** `[0, 1, 2, 3]`
- **MEX of the entire array:** 4 (since 0, 1, 2, and 3 are present, the smallest non-negative integer not in the array is 4).
- **Optimal Operation:** Set all elements to 4, resulting in the array `[4, 4, 4, 4]`.
- **Final Sum:** 16 (4 * 4).

This test case is designed to:
- Test the ability to compute the MEX correctly.
- Test the ability to apply the operation to the entire array.
- Ensure that the maximum sum is achieved with the correct number of operations.

By using this input, we can observe if the given code correctly computes the MEX, applies the operation, and outputs the correct result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
#include <bits/stdc++.h>
#define int long long
#define sqr(x) ((x) * (x))
#define MASK(i) (1LL << (i))
using namespace std;
typedef pair<int, int> pi;
const int mod = 998244353;

int32_t main() {
    #ifdef LOCAL
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t = 1;
//    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        vector<int> a(n+1);
        for(int i = 1; i <= n; i++)
            cin >> a[i];
        vector<vector<int>> max_sum(n+1, vector<int>(n+1));
        for(int i = 1; i <= n; i++) {
            for(int j = i; j <= n; j++) {
                if(i != j) {
                    max_sum[i][j] = (j-i+1) * (j-i+1);
                }
                else {
                    max_sum[i][j] = max(a[i], (j-i+1) * (j-i+1));
                }
            }
        }
        vector<pair<int, int>> dp(n+1);
        dp[0] = {0, -1};
        for(int i = 1; i <= n; i++) {
            for(int j = 0; j < i; j++) {
                dp[i] = max(dp[i], make_pair(dp[j].first + max_sum[j+1][i], j));
            }
        }
        cout << dp[n].first << " ";
        vector<pair<int, int>> ans;
        int r = n;
        function<void(int, int)> get_max = [&] (int l, int r) {
            if(a[l] != 0) {
                a[l] = 0;
                ans.emplace_back(l, l);
            }
            if(l < r) {
                get_max(l+1, r);
            }
            ans.emplace_back(l, r);
            for(int i = l; i <= r; i++) {
                a[i] = a[r] + 1;
            }
        };
        while(r > 0) {
            int l = dp[r].second + 1;
            if(max_sum[l][r] == sqr(r-l+1)) {
                for(int i = l; i <= r; i++) {
                    if(a[i] != 0) {
                        a[i] = 0;
                        ans.emplace_back(i, i);
                    }
                }
                get_max(l, r);
            }
            r = l - 1;
        }
        cout << ans.size() << "\n";
        for(int i = 0; i < ans.size(); i++) {
            cout << ans[i].first << " " << ans[i].second << "\n";
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5
0 1 2 3 4
```

### Explanation:
This test case is designed to challenge the buggy code by creating a scenario where the MEX operation needs to be applied in a specific way to achieve the maximum sum.

1. **Initial Array:**
   - The array is `[0, 1, 2, 3, 4]`.

2. **MEX Calculation:**
   - The MEX of the entire array `[0, 1, 2, 3, 4]` is `5` because `5` is the smallest non-negative integer not present in the array.
   - Applying the MEX operation to the entire array will set all elements to `5`.

3. **Expected Output:**
   - After the operation, the array becomes `[5, 5, 5, 5, 5]`.
   - The sum of the array is `5 * 5 = 25`.
   - The number of operations required is `1`.

4. **Buggy Code Analysis:**
   - The buggy code incorrectly calculates the maximum sum and the operations. It should recognize that applying the MEX operation to the entire array is the optimal solution.
   - The buggy code might not correctly handle the MEX operation or the dynamic programming part, leading to an incorrect result.

This test case should expose the bug in the buggy code by requiring a single operation that maximizes the sum of the array.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
#author : Sushmanth

from sys import stdin
input = stdin.readline

inp = lambda : list(map(int , input().split()))

from types import GeneratorType
def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc

def getmex(a):

    n = len(a)
    present = [False for i in range(n + 1)]
    for x in a:
        if(x > n):continue
        present[x] = True

    for i in range(n + 1):
        if(not present[i]):
            return i

@bootstrap
def solve(a , l , r):

    global bestopp

    n = len(a)
    if(getmex(a[l : r + 1]) == 0):
        bestopp.append([l + 1 , r + 1])
    else:
        bestopp.append([l + 1 , r + 1])
        bestopp.append([l + 1 , r + 1])

    for i in range(l , r + 1):
        a[i] = 0

    if(l == r):yield

    yield solve(a , l , r - 1)
    bestopp.append([l + 1 , r + 1])
    mex = getmex(a[l : r + 1])
    for i in range(l , r + 1):
        a[i] = mex

    yield solve(a , l , r - 1)

    yield

def answer():

    global bestopp

    max_mex = [[0 for i in range(n)] for j in range(n)]
    opp = [[None for i in range(n)] for j in range(n)]
    for l in range(n):
        for r in range(l , n):

            bestmex , bestopp = r - l + 1 , []
            b = a[:]
            solve(b , l , r)
            #print(b , l , r)

            max_mex[l][r] = bestmex
            opp[l][r] = list(bestopp)

    bestans , bestopp = 0 , []
    for mask in range(1 << n):

        i , ans , opperations = 0 , 0 , []
        while(i < n):
            key = (mask >> i & 1)
            j = i
            while(i < n and key == (mask >> i & 1)):
                if(key == 0):ans += a[i]
                i += 1

            if(key == 1):
                ans += (i - j) * max_mex[j][i - 1]
                opperations.extend(opp[j][i - 1])
                opperations.append([j + 1 , i])

        if(bestans < ans):
            bestans = ans
            bestopp = opperations[:]

    print(bestans , len(bestopp))
    for x in bestopp:
        print(*x)

for T in range(1):

    n = int(input())
    a = inp()

    answer()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
0 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
import sys
li = lambda: sys.stdin.readline().rstrip()
lli = lambda: sys.stdin.readline().rstrip().split()
Q = lambda: list(map(int, lli()))
from functools import lru_cache
from collections import Counter, defaultdict

n = int(li())
nums = Q()
# nums中出现0就一定替换它  并且 可以考虑向左向右 向前看...
# 任何元素都可以随时替换为0 从而去匹配我们所需的 1
def dp(a, b):
    s = nums[a:b+1]
    if 0 in s:
        k = set()
        t = 0
        for q,x in enumerate(s):
            k.add(x)
            while t in k: t += 1  # 可以替换的数更大了
        return t * (b-a+1), []
    else:
        # 0 不在其中 选择一个数变为0
        st = 0
        z = []
        for i in range(a, b+1):
            k = set([0])  # 长度为1 时t只能够为0
            t = 0
            for q,x in enumerate(s):
                if q+a == i: continue
                k.add(x)
                while t in k: t += 1  # 可以替换的数更大了
            if t * (b-a+1) > st:
                st = t * (b-a+1)
                z = [[i+1, i+1]]  # 选择它
        return st, z
@lru_cache(None)
def dfs(i):
    if i == n: return [0, []]
    d = [0, []]
    # 选
    for j in range(i, n):
        # 一直到 j 进行 mex 的替换
        a,b = dfs(j+1)
        c,g = dp(i, j)
        s = a + c
        if s > d[0]:
            if g:
                d = [s, b+g+[[i+1,j+1]]]
            else:
                d = [s, b+[[i+1,j+1]]]
    # 不选
    a,b = dfs(i+1)
    s = a + nums[i]
    if s > d[0]:
        d = [s, b] # 不加次数
    return d
# print(dfs(0))
a,b = dfs(0)
print(a, len(b))
for i in b:
    print(*i)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
0 0 0 0
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
//La illaha illalla muhammadur rasul Allah (SAW)
	//Bismillah hir-rehamn nir-rahim

	import java.io.*;
	import java.util.*;

	public class ComdeFormces {
		static int x[]= {0,-1,1,0,1,1,-1,-1};
		static int y[]= {-1,0,0,1,1,-1,1,-1};
		public static int m=(int)(1e9+7);
//		static HashMap<Integer,Integer> dp[][];
		static long ct;
		static long ans[];
		static long dp[];
		 static class Trie{
			    Trie a[];
				public Trie() {
					this.a=new Trie[2];
				}
			}
		 static ArrayList<pair> ar;

				public static void main(String[] args) throws Exception{
					// TODO Auto-generated method stub
			FastReader sc=new FastReader();
			BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
			//		 OutputStream out = new BufferedOutputStream ( System.out );
			int tc=1;
			int t=1;
			while(t--!=0) {
				int n=sc.nextInt();
				ar=new ArrayList<>();
				int a[]=new int[n];
				for(int j=0;j<n;j++) {
					a[j]=sc.nextInt();
				}
				int max=0;
				int ans=0;
				for(int i=0;i<(1<<n);i++) {
				   int cnt=0;
				   int sm=0;
				   for(int j=0;j<n;j++) {
					   int bt=i&(1<<j);
					   if(bt>0)cnt++;
					   else {
						   sm+=a[j]+cnt*cnt;
						   cnt=0;
					   }
				   }
				   sm+=cnt*cnt;
				   if(max<sm) {
					   max=sm;
					   ans=i;
				   }
				}
				int p=-1,q=-1;
				for(int j=0;j<n;j++) {
					int bt=(1<<j)&ans;
					if(bt>0) {
						if(p==-1)p=j;
						q=j;
					}
					else {
						if(p==-1)continue;
						else {
						  for(int k=p;k<=q;k++) {
							  if(a[k]>0)ar.add(new pair(k,k));
						  }
						  eval(p,q);
						}
						p=-1;q=-1;
					}
				}
				if(p!=-1) {
					  for(int k=p;k<=q;k++) {
						  if(a[k]>0)ar.add(new pair(k,k));
					  }
					  eval(p,q);
				}
				log.write(max+" "+ar.size()+"\n");
				for(pair el:ar)log.write((el.a+1)+" "+(el.b+1)+"\n");

			}
			log.flush();
		}
static void eval(int i,int j) {
	if(i==j) {
		ar.add(new pair(i,j));
		return;
	}
//	if(i==j+1) {
//		ar.add(new pair(i,i));
//		ar.add(new pair(i,j));
//		ar.add(new pair(i,i));
//		ar.add(new pair(i,i));
//		return;
//	}
	for(int k=j-1;k>=i;k--) {
		if(k==i) {
			ar.add(new pair(i,i));
			break;
		}
		eval(i,k-1);
		ar.add(new pair(i,k));
		ar.add(new pair(i,k-1));
	}
	ar.add(new pair(i,j));
}
	static void prime(int x,HashMap<Integer,Integer> hm) {
		int ct=0;
		while(x%2==0) {
			x/=2;
			ct++;
		}
		if(ct>0)hm.put(2, hm.getOrDefault(2, 0)+ct);
		for(int i=3;i*i<=x;i+=2) {
			ct=0;
			while(x%i==0) {
				x/=i;
				ct++;
			}
			if(ct>0)hm.put(i, hm.getOrDefault(i, 0)+ct);
		}
		if(x>1) {
			hm.put(x, hm.getOrDefault(x, 0)+1);
		}
	}
	static void add(TreeMap<Integer,Integer> tm,int el) {
		tm.put(el, tm.getOrDefault(el, 0)+1);
	}
	static void rem(TreeMap<Integer,Integer> tm,int el) {
		tm.put(el, tm.get(el)-1);
		if(tm.get(el)==0)tm.remove(el);
	}
	// small function bazi due to alas
	static int max(int a,int b) {return a>=b?a:b;}
	static int min(int a,int b) {return a>=b?b:a;}
	static long max(long a,long b) {return a>=b?a:b;}
	static long min(long a,long b) {return a>=b?b:a;}
	static int abs(int a) {
		return a>=0?a:-a;
	}

	static long abs(long a) {
	return a>=0?a:-a;
	}

	static long div(long a,long b) {
		long an=pow(b,m-2)%m;
		return mul(an,a);
	}
	public static long mul(long a, long b) {
		return (a*b)%m;
	}
	public static long add(long a, long b) {
		return (a+b)%m;
	}
	public static long sub(long a,long b) {
	  return (a%m-b%m+m)%m;
	}
	static int gcd(int a,int b) {
		if(b==0)return a;
		else return gcd(b,a%b);
	}
	static long gcd(long a,long b) {
		if(a==0)return 0;
		if(b<2*a)return 1+gcd(b,abs(a-b));
		long val=b/(2*a);
		return 3*val+gcd(a,b-2*a*val);
	}
	static long ncr(int n, int r){
	    if(r>n-r)r=n-r;
	    long ans=1;
	    for(int i=0;i<r;i++){
	          ans=mul(ans,(n-i));
	          ans=div(ans,i+1);
	    }
	    return ans;
	}
	public static class trip{
		int a;
		int b;
		int c;
		public trip(int a,int b,int c) {
			this.a=a;
			this.b=b;
			this.c=c;
		}
	}
	static void mergesort(int[] a,int start,int end) {
			if(start>=end)return ;
			int mid=start+(end-start)/2;
			mergesort(a,start,mid);
			mergesort(a,mid+1,end);
			merge(a,start,mid,end);

		}
	static void merge(int[] a, int start,int mid,int end) {
			int ptr1=start;
			int ptr2=mid+1;
			int b[]=new int[end-start+1];
			int i=0;
			while(ptr1<=mid && ptr2<=end) {
				if(a[ptr1]<=a[ptr2]) {
					b[i]=a[ptr1];
					ptr1++;
					i++;
				}
				else {
					b[i]=a[ptr2];
					ptr2++;
					i++;
				}
			}
			while(ptr1<=mid) {
				b[i]=a[ptr1];
				ptr1++;
				i++;
			}
			while(ptr2<=end) {
				b[i]=a[ptr2];
				ptr2++;
				i++;
			}
			for(int j=start;j<=end;j++) {
				a[j]=b[j-start];
			}
		}
	 public static class FastReader {

			BufferedReader b;
			StringTokenizer s;
			public FastReader() {
				b=new BufferedReader(new InputStreamReader(System.in));
			}
			String next() {
				while(s==null ||!s.hasMoreElements()) {
					try {
						s=new StringTokenizer(b.readLine());
					}
					catch(IOException e) {
						e.printStackTrace();
					}
				}
				return s.nextToken();
			}
			public int nextInt() {
				return Integer.parseInt(next());
			}
			public long nextLong() {
				return Long.parseLong(next());
			}
			public double nextDouble() {
				return Double.parseDouble(next());
			}
			String nextLine() {
				String str="";
				try {
					str=b.readLine();
				}
				catch(IOException e) {
					e.printStackTrace();
				}
				return str;
			}
			boolean hasNext() {
			    if (s != null && s.hasMoreTokens()) {
			        return true;
			    }
			    String tmp;
			    try {
			        b.mark(1000);
			        tmp = b.readLine();
			        if (tmp == null) {
			            return false;
			        }
			        b.reset();
			    } catch (IOException e) {
			        return false;
			    }
			    return true;
			}

	}
	public static class pair{
		int a,b,c,d,e;
		public pair(int a,int b,int c) {
			this.a=a;
			this.b=b;
			this.c=c;
		}
		public pair(int a,int b) {
			this.a=a;
			this.b=b;
		}
		public pair(int a,int b,int c,int d) {
			this.a=a;
			this.b=b;
			this.c=c;
			this.d=d;
		}
		public pair(int a,int b,int c,int d,int e) {
			this.a=a;
			this.b=b;
			this.c=c;
			this.d=d;
			this.e=e;
		}
	//			@Override
	//			public String toString() {
	//				return "{"+this.a+" "+this.b+"}";
	//				}
		}
			static long pow(long a, long pw) {
				long temp;
				if(pw==0)return 1;
				temp=pow(a,pw/2);
				if(pw%2==0)return mul(temp,temp);
				return mul(a,mul(temp,temp));

			}

			static long poww(long a, long pw) {
				long temp;
				if(pw==0)return 1;
				temp=poww(a,pw/2);
				if(pw%2==0)return temp*temp;
				return a*temp*temp;

			}
			static int poww(int a, int pw) {
				int temp;
				if(pw==0)return 1;
				temp=poww(a,pw/2);
				if(pw%2==0)return temp*temp;
				return a*temp*temp;

			}

	}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input for the given buggy code, we need to focus on edge cases that can expose potential issues in the algorithm. Here are some considerations:

1. **Edge Cases:**
   - Arrays with all elements being the same.
   - Arrays with a mix of large and small values.
   - Arrays with elements that are all small or all large.
   - Arrays with elements that are just below the MEX value to ensure the MEX operation is correctly handled.

2. **Small-Scale Test Input:**
   - We need to ensure the input is small enough to fit within the constraints but challenging enough to expose bugs.

3. **Challenging Input:**
   - An array that has a mix of values and a specific pattern that can cause the algorithm to fail in computing the MEX correctly or in performing the operations optimally.

Given these considerations, a challenging test input could be:

```plaintext
4
1 2 3 0
```

### Explanation:
- **Array: [1, 2, 3, 0]**
  - The MEX of the entire array is 4.
  - The optimal solution would be to set the entire array to 4, resulting in the array [4, 4, 4, 4] and a sum of 16.
  - This input tests the ability of the algorithm to correctly compute the MEX and apply the operation to maximize the sum.

This input is small and complete, and it should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
import java.util.TreeSet;

public class D_1956 {
    public static int solve(int start, int end, int[]arr, ArrayList<int[]> changeSub){
        if(start > end){
            return 0;
        }
        if(start == end){
            if(arr[start] == 0){
                changeSub.add(new int[]{start, start});
                return 1;
            }
            return arr[start];
        }

        int max = arr[start];
        int index = start;

        int total = 0;
        for(int i = start ; i <= end ; i++){
            if(arr[i] > max){
                max = arr[i];
                index = i;
            }
            total += arr[i];
        }

        int temp = (end - start + 1) * (end - start + 1);
        if(temp > total){
            changeSub.add(new int[]{start, end});
            return temp;
        }

        return solve(start, index-1, arr, changeSub) + arr[index] + solve(index+1, end, arr, changeSub);
    }

    public static void findSub(int start, int end, int []arr, ArrayList<int []> ansList){
        if(start > end){
            return ;
        }
        ArrayList<int []> ans1 = new ArrayList<>();
        ArrayList<int []> ans2 = new ArrayList<>();

        ArrayList<int[]> list = new ArrayList<>();
        for(int i = 0 ; i <= end-start ; i++){
            ArrayList<int[]> temp = new ArrayList<>(list);
            list.addAll(temp);
            list.add(new int[]{start,start+i});

            if(arr[i+start] != i){
                ans1.addAll(list);
            }
        }

        list = new ArrayList<>();
        for(int i = end-start ; i >= 0 ; i--){
            ArrayList<int[]> temp = new ArrayList<>(list);
            list.addAll(temp);
            list.add(new int[]{start+i,end});

            if(arr[i+start] != i){
                ans2.addAll(list);
            }
        }
        if(ans2.size() > ans1.size())
            ansList.addAll(ans1);
        else
            ansList.addAll(ans2);
    }

    public static void addMinRange(int start, int end, int []arr, ArrayList<int[]> ansList){
        int[] newArr = new int[end-start+1];
        for(int i = 0 ; i < end-start+1 ; i++){
            newArr[i] = arr[i+start];
        }

        int n = newArr.length;

        int start1 = 0;
        int end1 = n-1;

        int start2 = 0;
        int end2 = n-1;

        ArrayList<Integer> ts = new ArrayList<>();

        int max = n-1;
        for(int i = 0 ; i < n ; i++){
            if(ts.contains(newArr[i]))
                break;
            ts.add(newArr[i]);
            Collections.sort(ts);
            if(ts.get(i) == n-1 && ts.get(0) == n-1-i){
                start1 = i+1;
            }
        }

        ts = new ArrayList<>();
        max = n-1-start1;
        for(int i = n-1 ; i >= 0 ; i--){
            if(newArr[i] > max || ts.contains(newArr[i]))
                break;
            ts.add(newArr[i]);
            Collections.sort(ts);
            if(ts.get(n-1-i) == max && ts.get(0) == max-(n-1-i)){
                end1 = i-1;
            }
        }

        ts = new ArrayList<>();
        max = n-1;
        for(int i = n-1 ; i >= 0 ; i--){
            if(ts.contains(newArr[i]))
                break;
            ts.add(newArr[i]);
            Collections.sort(ts);
            if(ts.get(n-1-i) == max && ts.get(0) == max-(n-1-i)){
                end2 = i-1;
            }
        }

        ts = new ArrayList<>();
        max = end2;

        for(int i = 0 ; i < n ; i++){
            if(newArr[i] > max || ts.contains(newArr[0]))
                break;
            ts.add(newArr[i]);
            Collections.sort(ts);
            if(ts.get(i) == n-1 && ts.get(0) == n-1-i){
                start2 = i+1;
            }
        }

        if(end1-start1 < end2-start2){
//            System.out.println((start1+start) + " " + (end1+start));
            findSub(start1+start, end1+start, arr, ansList);
        }
        else{
//            System.out.println((start2+start) + " " + (end2+start));
            findSub(start2+start, end2+start, arr, ansList);
        }

        ansList.add(new int[]{start, end});
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int []arr = new int[n];

        for(int i = 0 ; i < n ; i++){
            arr[i] = sc.nextInt();
        }
        ArrayList<int[]> changeSub = new ArrayList<>();
        int ans = solve(0, n-1, arr, changeSub);
        ArrayList<int []> ansList = new ArrayList<>();
        for(int i = 0 ; i < changeSub.size() ;i++){
            addMinRange(changeSub.get(i)[0], changeSub.get(i)[1], arr, ansList);
        }
        System.out.println(ans + " " + ansList.size());
        for (int[] x:ansList) {System.out.println((x[0]+1) + " " + (x[1]+1));}
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
0 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define ll long long
#define pll pair<ll, ll>
#define f first
#define s second
#define mod 1000000007
#define mid (l+r)/2
#define ld long double
using namespace std;
using namespace __gnu_pbds;
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update>ordered_set;
ll phi(ll n){
    ll result=n;
    for(ll p=2; p*p<=n; ++p){
        if(n%p==0){
            while(n%p==0)n/=p;
            result-=result/p;
        }
    }
    if(n>1)result-=result/n;
    return result;
}
ll binpow(ll a, ll b){
    if(b==0)return 1;
    if(b==1)return a;
    ll num=binpow(a, b>>1)%mod, p=1;
    if(b%2)p=a;
    ll part=(num*num)%mod;
    return(part*p)%mod;
}
ll eucl(pll a, pll b){
    return (a.first-b.first)*(a.first-b.first)+(a.second-b.second)*(a.second-b.second);
}
void multi(vector<vector<ll>>&a, vector<vector<ll>>&b){
    vector<vector<ll>>c(3, vector<ll>(3, 0));
    for(int i=0; i<3; i++)for(int j=0; j<3; j++)for(int k=0; k<3; k++)c[i][j]+=(a[i][k]*b[k][j]);
    for(int i=0; i<3; i++)for(int j=0; j<3; j++)a[i][j]=c[i][j];
}
vector<vector<ll>>power(vector<vector<ll>>M, ll expo){
    vector<vector<ll>>iden={{1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
    while(expo){
        if(expo&1)multi(iden, M);
        multi(M, M);
        expo>>=1;
    }
    return iden;
}
ll matex(ll n){
    if(n==0)return 0;
    if(n==1)return 1;
    if(n==2)return 2;
    vector<vector<ll>>M={{1, 1, 0}, {0, 1, 1}, {0, 0, 1}};
    vector<vector<ll>>S={{2, 0, 0}, {3, 0, 0}, {1, 0, 0}};
    vector<vector<ll>>res=power(M, n-3);
    multi(res, S);
    return res[0][0];
}
const int movx[]={-1, 0, 1, 0}, movy[]={0, 1, 0, -1};
vector<pll>recc;
void edit(vector<ll>&v, ll idx, ll ini){
    if(idx==ini){
        if(v[idx]!=0){
            v[idx]=0;
            recc.push_back({idx, idx});
            }
        return;
    }
    for(int i=idx-1; i>=ini; i--){
        if(v[i]!=i-ini){
        edit(v, i, ini);
        if(ini!=i)recc.push_back({ini, i});
        }
        for(int j=ini; j<=i; j++)v[j]=i-ini;
    }
    v[idx]=idx-ini;
    for(int j=ini; j<=idx; j++)v[j]=idx-ini;
}
int main(){
        ios_base::sync_with_stdio(0);
        cin.tie(0); cout.tie(0);
        ll t=1;
        //cin>>t;
        while(t--){
            ll n;
            cin>>n;
            vector<ll>v(n+1);

            for(int i=1; i<=n; i++)cin>>v[i];
            for(int p=1; p<=100; p++){
            for(int i=1; i<=n; i++){
                ll sum=0, idx=0;
                for(int j=i; j<=n; j++){
                    sum+=v[j];
                    if(sum<(j-i+1)*(j-i+1)){
                        idx=j;
                        sum=(j-i+1)*(j-i+1);
                    }
                }
                if(idx>=i){
                edit(v, idx, i);
                for(int j=i; j<=idx; j++)v[j]=idx-i+1;
                recc.push_back({i, idx});
                }
            }
            }
            ll sum=0;
            for(int i=1; i<=n; i++)sum+=v[i];
            cout<<sum<<" "<<recc.size()<<'\n';
            for(auto u:recc)cout<<u.first<<" "<<u.second<<'\n';
        }
        return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
import typing
import sys
from collections import defaultdict
input = lambda: sys.stdin.readline().strip()
inf = 10**18
mod = 998244353
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')
# sys.setrecursionlimit(10**6)

def solve():
  N = int(input())
  A = list(map(int, input().split()))
  stay = [False for _ in range(N)]
  ops = []
  # [0, 1,..., ln-1]
  def op_s(l, ln):
    if ln == 1:
      # ops.append((l, l))
      return
    op_s(l, ln-1)
    ops.append((l, l+ln-1)) # 全体
    ops.append((l, l+ln-2)) # 消す
    op_s(l, ln-1) # 消す

  # 全て0の状態から[ln] * lnにする
  def op_(l, ln):
    for i in range(ln):
      # i->i+1
      # [i,..i,0,0]->[i+1,..i+1,i+1,0]
      # [i,..,i,i,0,0]->[0,..,0,i,0,0] # iがi+1個あるところからi個削る
      if i == 0:
        ops.append((l, l+min(1, ln-1)))
      if i >= 1:
        ops.append((l, l+i-1)) # i以外を0にする
        op_s(l, i) # i個の階段 [0, 1, ..., i-1,...
        ops.append((l, l+i)) # i+1にする

    # この時点で [ln-1]+ln-1

  # 始点をlとして，長さlnの配列を全てlnにする
  def op(l, ln, arr):
    for i in range(l, l+ln):
      # 一旦全て0にする
      if arr[i] != 0:
        ops.append((i,i))
    op_(l, ln)

  def dfs(arr, l, r):
    ln = r-l
    # 区間0は無視
    if ln == 0:
      return
    flag = False
    cur = l-1
    for i in range(l, r):
      if A[i] > ln:
        dfs(arr, cur+1, i)
        cur = i
        flag = True
    # 最後のセクション
    # このとき必ずflag = True
    if cur >= l:
      dfs(arr, cur+1, r)

    # 再帰で潜った場合
    if flag:
      return
    # 全ての要素がln以下のとき
    # 操作を出力
    op(l, ln, arr)
    # 全てをlnにする
    for i in range(l, r):
      arr[i] = ln

  dfs(A, 0, N)
  print(sum(A), len(ops))
  for l, r in ops:
    print(l+1, r+1)
def main():
  t = 1
  for _ in range(t):
    solve()
main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
0 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
import sys
sys.setrecursionlimit(int(2e5))
input = sys.stdin.readline
def li(): return list(map(int,input().split()))
def n_k(): return map(int,input().split())
def mexi(l,r):
    P = set()
    for k in range(l,r):
        P.add(lis[k])
    MM = 0
    for k in range(r-l):
        if k in P: MM+=1
        else:break
    for k in range(l,r):
        lis[k] = MM
SOL = []
def pp(l,r,F):
    h0 = False

    for i in range(l,r):
        if lis[i] == 0:
            h0 = True
    SOL.append([l,r]);mexi(l,r)

    if h0 :SOL.append([l,r]);mexi(l,r)

    for i in range(1,r-l):
        pp(l,r-i,1-F)
    SOL.append([l,r]);mexi(l,r)
for _ in range(1):
    n = int(input())
    lis = li()
    M = 0
    MI = 0
    for i in range(2**n):
        ff = i
        SC=0
        ST=0
        for j in range(n):
            if ff % 2:
                ST+=1
            else:
                SC+=ST**2
                SC+=lis[j]
                ST=0
            ff//=2
        SC+=ST**2
        if SC > M:
            MI=i
            M=SC

    ST = 0
    for j in range(n):
        if MI % 2:
            ST += 1
        else:
            if ST:
                pp(j-ST,j)
                ST=0
        MI //= 2
    if ST:
        pp(n-ST,n,1)
print(M,len(SOL))
for a,b in SOL:
    print(a+1,b)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
1 2 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {
boolean MULTI_CASE = false;
boolean ALWAYS_FLUSH = false;

void go() {
    int n = nextInt();
    a = new int[n];
    for (int i = 0; i < a.length; ++i) a[i] = nextInt();
    int ret = 0, retT = -1;
    for (int t = 0; t < (1 << n); ++t) {
        int sum = 0;
        for (int i = 0; i < n; ++i) {
            if ((t & (1 << i)) > 0) {
                int j = i;
                while (j < n && (t & (1 << j)) > 0) j++;
                sum += (j - i) * (j - i);
                i = j - 1;
            } else sum += a[i];
        }
        if (sum > ret) {
            ret = sum;
            retT = t;
        }
    }
    pn = 0;
    for (int i = 0; i < n; ++i) {
        if ((retT & (1 << i)) > 0) {
            int j = i;
            while (j < n && (retT & (1 << j)) > 0) j++;
            // i .. j-1
            int l = i + 1, r = j;
            proc(l, r);
            i = j - 1;
        }
    }
    sl(ret + " " + pn);
    for (int i = 0; i < pn; ++i) {
        sl(p[i][0] + " " + p[i][1]);
    }
}

int[] a;
int[][] p = new int[500500][2];
int pn;

void proc(int l, int r) {
    for (int i = l; i <= r; ++i) {
        if (a[i - 1] != 0) add(i, i);
    }
    int len = r - l + 1;
    get(len - 1, l);
    add(l, r);
}

// get num, num-1, num-2, ..., 0 starting at position l.
// from 0, 0, 0, ..., 0.
void get(int num, int l) {
    if (num == 0) return;
    add(l, l);
    if (num == 1) return;
    get(num - 1, l);
    add(l, l + num - 1);
    for (int i = 1; i <= num - 1; i++) {
        add(l + i, l + i);
    }
    get(num - 1, l + 1);
}

void add(int x, int y) {
    p[pn++] = new int[]{x, y};
}

/**
 * Input.
 */
InputStream inStream;
byte[] inBuff = new byte[1024];
int inBuffCursor = 0, inBuffLen = 0;

boolean isVisibleChar(int c) {
    return 33 <= c && c <= 126;
}

boolean hasNextByte() {
    if (inBuffCursor < inBuffLen) return true;
    inBuffCursor = 0;
    try {
        inBuffLen = inStream.read(inBuff);
    } catch (Exception e) {
        e.printStackTrace();
    }
    return inBuffLen > 0;
}

boolean hasNext() {
    while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;
    return hasNextByte();
}

int nextByte() {
    return hasNextByte() ? inBuff[inBuffCursor++] : -1;
}

String next() {
    if (!hasNext()) throw new RuntimeException("no next.");
    StringBuilder sb = new StringBuilder();
    int b = nextByte();
    while (isVisibleChar(b)) {
        sb.appendCodePoint(b);
        b = nextByte();
    }
    return sb.toString();
}

long nextLong() {
    if (!hasNext()) throw new RuntimeException("no next.");
    long result = 0;
    boolean negative = false;
    int b = nextByte();
    if (b < '0') {
        if (b == '-') negative = true;
        else if (b != '+') throw new RuntimeException("long number must start with +/-.");
        b = nextByte();
    }
    while (isVisibleChar(b)) {
        if (b < '0' || b > '9') throw new RuntimeException("wrong digit in long:" + (char) b);
        // TODO: may overflow here, even if it is a valid Long, eg.-(1<<63).
        result = result * 10 + (b - '0');
        b = nextByte();
    }
    return negative ? -result : result;
}

int nextInt() {
    long x = nextLong();
    if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)
        throw new RuntimeException("int overflow:" + x);
    return (int) x;
}

double nextDouble() {
    return Double.parseDouble(next());
}

/**
 * Output.
 */
PrintWriter printOut = new PrintWriter(System.out);

void so(Object obj, boolean newLine) {
    if (newLine) printOut.println(obj);
    else printOut.print(obj);
    if (ALWAYS_FLUSH) printOut.flush();
}

void so(Object obj) {
    so(obj, false);
}

void sl(Object obj) {
    so(obj, true);
}

void sl() {
    sl("");
}

/**
 * Main.
 */
void mainGo() {
    try {
        inStream = new FileInputStream("src/main.in");
        ALWAYS_FLUSH = true;
    } catch (Exception e) {
        inStream = System.in;
    }
    while (hasNext()) {
        if (MULTI_CASE) {
            int T = nextInt();
            if (T == 0) break;
            for (int i = 0; i < T; ++i) {
                go();
            }
        } else {
            go();
        }
    }
    printOut.flush();
}

public static void main(String[] args) throws Exception {
    new Main().mainGo();
}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
0 1 2 3
```

### Explanation:
This test case is designed to expose the bug in the provided code by focusing on a small array where the MEX operation can significantly alter the array values.

- **Initial Array:** `[0, 1, 2, 3]`
- **Expected Outcome:** The maximum possible sum is achieved by setting all elements to 4, as the MEX of the entire array is 4. The final array should be `[4, 4, 4, 4]`, with a sum of 16.

### Steps to Verify:
1. **Initial Sum:** The initial sum of the array is 6.
2. **MEX Calculation:** The MEX of the entire array `[0, 1, 2, 3]` is 4.
3. **Operation:** Perform the operation on the entire array, setting all elements to 4.
4. **Final Sum:** The final sum is 16.

### Why This Test Case:
- **Small Scale:** The array length is 4, which is within the problem constraints and manageable for manual verification.
- **Simple MEX:** The MEX of the array is straightforward to calculate and should be 4.
- **Maximization:** The operation should clearly maximize the sum, making it easy to spot if the code fails to achieve the correct result.

This test case should help identify if the code correctly computes the MEX and applies the operation to maximize the sum.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

/**
 *
 * @author is2ac
 */
public class A_CF {

    static final Random RAND = new Random();
   static int ct = 0;
   static StringBuilder sb;
    public static void main(String[] args) {
        FastScanner55 fs = new FastScanner55();
        PrintWriter pw = new PrintWriter(System.out);
        //int t = fs.ni();
        int t = 1;
        //pw.println("");
        sb =new StringBuilder();
        for (int tc = 1; tc <= t; tc++) {
            sb = new StringBuilder();
            int n = fs.ni();
            int[] a = fs.intArray(n);
            Integer[] dp = new Integer[n];
            ct = 0;
            int res = recur(0,a,dp);
            trace(0,a,dp);
            pw.println(res + " " + ct);
            pw.println(sb);
            //pw.println(sb2);
        }

        pw.close();

    }
    public static void trace(int ind, int[] a, Integer[] dp) {
        if (ind==a.length) return;
        //System.out.println("at idn " + ind);
        int cur = a[ind]+recur(ind+1,a,dp);
        if (cur==dp[ind]) {
            trace(ind+1,a,dp);
            return;
        }
        for (int i = ind; i < a.length ; i++) {
            cur = (int)Math.pow(i-ind+1,2)+recur(i+1,a,dp);
            if (cur==dp[ind]) {
                int n = i-ind+1;
                ops(ind,i,a,n-1);
                ct++;
                sb.append((1+ind) + " " + (1+i));
                sb.append("\n");
                for (int j = ind; j <= i; j++) {
                }
                trace(i+1,a,dp);
                return;
            }
        }
    }
    public static void ops(int l, int r, int[] a, int h) {
        for (int i = l; i <= r; i++) {
            if (a[i]!=0) {
                ct++;
                sb.append((1+i) + " " + (1+i));
                sb.append("\n");
                a[i] = 0;
            }
        }
        if (h==1) {
            ct++;
                sb.append((1+r) + " " + (1+r));
                sb.append("\n");
                a[r] = 1;
                return;
        }
        if (h==0) return;
        for (int i = 1; h-i>=0; i++) {
             ops(l,r-i+1,a,h-i);
        }
        ct++;
        sb.append((1+l) + " " + (1+r));
        sb.append("\n");
        for (int i = l; i <= r; i++) {
                a[i] = h;
        }
        ops(l,r-1,a,h-1);
        //ops(l,r-1,a,h-1);
    }
    public static int recur(int ind, int[] a, Integer[] dp) {
        if (ind==a.length) return 0;
        if (dp[ind]!=null) return dp[ind];
        int res = recur(ind+1,a,dp) + a[ind];
        for (int i = ind; i < a.length;i ++) {
            res = Math.max(res,(int)Math.pow(i-ind+1,2)+recur(i+1,a,dp));
        }
        return dp[ind] = res;
    }
    public static int gcd(int a, int b) {
        if (b==0) return a;
        return gcd(b,a%b);
     }

    public static void sort(int[] arr) {
        for (int i = 0; i < arr.length; i++) {
            int r = RAND.nextInt(arr.length);
            int temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
}

class FastScanner55 {

    BufferedReader br;
    StringTokenizer st;

    public FastScanner55() {
        br = new BufferedReader(new InputStreamReader(System.in), 32768);
        st = null;
    }

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    Integer nexti () {
        if (st == null || !st.hasMoreElements()) return null;
        return Integer.parseInt(next());
    }
    int ni() {
        return Integer.parseInt(next());
    }

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++) {
            ret[i] = ni();
        }
        return ret;
    }

    long nl() {
        return Long.parseLong(next());
    }

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++) {
            ret[i] = nl();
        }
        return ret;
    }

    double nd() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following aspects:

1. **Edge Cases**: Small arrays, arrays with all elements the same, arrays with all elements different, and arrays with specific values that might trigger incorrect MEX calculations.
2. **Boundary Conditions**: Arrays of length 1, 2, and 18.
3. **Complex Scenarios**: Arrays that require multiple operations to maximize the sum, especially where the MEX calculation is non-trivial.

Given the constraints and the need for a small-scale test input, let's focus on an array of length 18, which is the upper limit. We will create an array that has a mix of values to ensure the MEX calculation is not straightforward and requires careful handling.

### Challenging Test Input
```plaintext
18
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
```

### Explanation
- **Array Length**: 18, the maximum allowed length.
- **Array Values**: A sequence from 0 to 17, ensuring that the MEX calculation is non-trivial and requires multiple operations to maximize the sum.
- **MEX Calculation**: The MEX of the entire array is 18, but the MEX of smaller subarrays will vary, making it a good test for the algorithm's ability to handle different MEX values and perform the operations correctly.

This test input should expose any issues in the buggy code related to MEX calculation and the sequence of operations to maximize the sum.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Nene and the Mex Operator
time_limit_ms: 2000
memory_limit_mb: 256
Description: Nene gave you an array of integers $$$a_1, a_2, \ldots, a_n$$$ of length $$$n$$$.

You can perform the following operation no more than $$$5\cdot 10^5$$$ times (possibly zero):

- Choose two integers $$$l$$$ and $$$r$$$ such that $$$1 \le l \le r \le n$$$, compute $$$x$$$ as $$$\operatorname{MEX}(\{a_l, a_{l+1}, \ldots, a_r\})$$$, and simultaneously set $$$a_l:=x, a_{l+1}:=x, \ldots, a_r:=x$$$.

Here, $$$\operatorname{MEX}$$$ of a set of integers $$$\{c_1, c_2, \ldots, c_k\}$$$ is defined as the smallest non-negative integer $$$m$$$ which does not occur in the set $$$c$$$.

Your goal is to maximize the sum of the elements of the array $$$a$$$. Find the maximum sum and construct a sequence of operations that achieves this sum. Note that you don't need to minimize the number of operations in this sequence, you only should use no more than $$$5\cdot 10^5$$$ operations in your solution.

Input: The first line contains an integer $$$n$$$ ($$$1 \le n \le 18$$$) — the length of the array $$$a$$$.

The second line contains $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ ($$$0\leq a_i \leq 10^7$$$) — the array $$$a$$$.

Output: In the first line, output two integers $$$s$$$ and $$$m$$$ ($$$0\le m\le 5\cdot 10^5$$$) — the maximum sum of elements of the array $$$a$$$ and the number of operations in your solution.

In the $$$i$$$-th of the following $$$m$$$ lines, output two integers $$$l$$$ and $$$r$$$ ($$$1 \le l \le r \le n$$$), representing the parameters of the $$$i$$$-th operation.

It can be shown that the maximum sum of elements of the array $$$a$$$ can always be obtained in no more than $$$5 \cdot 10^5$$$ operations.

Examples:
input:
2
0 1
output:
4 1
1 2

input:
3
1 3 9
output:
13 0

input:
4
1 100 2 1
output:
105 2
3 3
3 4

input:
1
0
output:
1 1
1 1

Note: In the first example, after the operation with $$$l=1$$$ and $$$r=2$$$ the array $$$a$$$ becomes equal to $$$[2,2]$$$. It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$4$$$.

In the second example, the initial sum of elements is $$$13$$$ which can be shown to be the largest.

In the third example, the array $$$a$$$ changes as follows:

- after the first operation ($$$l=3$$$, $$$r=3$$$), the array $$$a$$$ becomes equal to $$$[1,100,0,1]$$$;
- after the second operation ($$$l=3$$$, $$$r=4$$$), the array $$$a$$$ becomes equal to $$$[1,100,2,2]$$$.

It can be shown that it is impossible to achieve a larger sum of the elements of $$$a$$$, so the answer is $$$105$$$.

Buggy Code:
#include <iostream>
#include <cmath>
#include <string>
#include <vector>
#include <algorithm>
#include <iomanip>
#include <queue>
#include <set>
#include <fstream>
#include <map>
#include <regex>
#include <random>
#include <time.h>
#include <climits>
using namespace std;
long long mod = 998244353;

vector<pair<long long, long long>> m;

void f(long long size, long long i, const long long orgsize)
{
    if(size == 1)
    {
        m.push_back({i + 1, i + 1});
        return;
    }

    f(size - 1, i, orgsize);
    m.push_back({i + 1, i + size});

    if(size == orgsize)
    {
        f(size - 1, i, orgsize);
        m.push_back({i + 1, i + size});
    }
    else
    {
        f(size - 1, i, orgsize);
    }

    return;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ifstream fin("input.txt");
    ofstream fout("output.txt");
    long long n;
    cin >> n;
    vector<long long> v(n + 1);
    for(int i = 1; i <= n; i++)
    {
        cin >> v[i];
    }

    if(n == 1 && v[1] == 0)
    {
        cout << "1 1" << endl;
        cout << "1 1" << endl;
        return 0;
    }

    vector<long long> dp(n + 1);
    dp[1] = v[1] == 0 ? 1 : v[1];
    for(int i = 2; i <= n; i++)
    {
        dp[i] = dp[i - 1] + v[i];
        for(int j = 0; j < i; j++)
        {
            dp[i] = max(dp[i], dp[j] + (i - j) * (i - j));
        }
    }

    vector<pair<long long, long long>> ans;
    long long i = n;
    while(i != 0)
    {
        if(dp[i] == dp[i - 1] + v[i])
        {
            i -= 1;
            continue;
        }

        for(int j = 0; j < i; j++)
        {
            if(dp[i] == dp[j] + (i - j) * (i - j))
            {
                ans.push_back({j, i - 1});
                i = j;
            }
        }
    }

    for(auto [i, j] : ans)
    {
        for(int k = i; k <= j; k++)
        {
            if(v[k + 1] == k - i)
                m.push_back({k + 1, k + 1});
        }

        f(j - i + 1, i, j - i + 1);
    }

    cout << dp[n] << " " << m.size() << endl;
    for(auto [i, j] : m)
    {
        cout << i << " " << j << endl;
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
0 1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
#include <iostream>
#include <string>

int main() {
    int n, m;
    std::cin >> n >> m;

    std::string s, t;
    std::cin >> s >> t;

    int i = 0, j = 0;
    int countS = 0, countT = 0;

    // Iterate through both strings simultaneously
    while (i < n && j < m) {
        // If the current characters match or one of them is '-', continue to the next character
        if (s[i] == t[j] || s[i] == '-' || t[j] == '-') {
            i++;
            j++;
        }
        // If the current character in s is '*', increment the count of '*' in s
        else if (s[i] == '*') {
            countS++;
            i++;
        }
        // If the current character in t is '*', increment the count of '*' in t
        else if (t[j] == '*') {
            countT++;
            j++;
        }
        // If the current characters do not match and neither of them is '*' or '-', return "No"
        else {
            std::cout << "No" << std::endl;
            return 0;
        }
    }

    // If there are remaining characters in s, add their count of '*' to countS
    while (i < n) {
        if (s[i] == '*') {
            countS++;
        }
        i++;
    }

    // If there are remaining characters in t, add their count of '*' to countT
    while (j < m) {
        if (t[j] == '*') {
            countT++;
        }
        j++;
    }

    // If the counts of '*' in both strings are equal, return "Yes". Otherwise, return "No".
    if (countS == countT) {
        std::cout << "Yes" << std::endl;
    } else {
        std::cout << "No" << std::endl;
    }

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
7 7
a*b*c*a
a*b*c*a
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
#include <iostream>
#include <string>

using namespace std;

bool wildcardMatch(string s, string p) {
    int i = 0, j = 0, star = -1, match;
    while (i < s.length()) {
        if (j < p.length() && (p[j] == '?' || p[j] == s[i])) {
            i++;
            j++;
        } else if (j < p.length() && p[j] == '*') {
            star = j;
            match = i;
            j++;
        } else if (star != -1) {
            j = star + 1;
            match++;
            i = match;
        } else {
            return false;
        }
    }
    while (j < p.length() && p[j] == '*') j++;
    return j == p.length();
}

int main() {
    int n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;
    cout << (wildcardMatch(s, t) ? "Yes" : "No") << endl;
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
6 6
a-b*c
a*b-c
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
#include <iostream>
#include <string>

using namespace std;

bool wildcardMatch(string s, string p) {
    int i = 0, j = 0, star = -1, match;
    while (i < s.length()) {
        if (j < p.length() && (p[j] == '?' || p[j] == s[i])) {
            i++;
            j++;
        } else if (j < p.length() && p[j] == '*') {
            star = j;
            match = i;
            j++;
        } else if (star != -1) {
            j = star + 1;
            match++;
            i = match;
        } else {
            return false;
        }
    }
    while (j < p.length() && p[j] == '*') j++;
    return j == p.length();
}

int main() {
    int n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;
    cout << (wildcardMatch(s, t) ? "Yes" : "No") << endl;
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5 5
a*b*c
a-b-c
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Zimpha Fan Club
time_limit_ms: 12000
memory_limit_mb: 512
Description: One day, Zimpha casually came up with a problem. As a member of "Zimpha fan club", you decided to solve that problem.      You are given two strings $s$ and $t$ of length $n$ and $m$, respectively. Both strings only consist of lowercase English letters, - and *.

You need to replace all occurrences of * and -, observing the following rules:

  * For each -, you must replace it with any lowercase English letter.    * For each *, you must replace it with a string of any (possibly, zero) length which only consists of lowercase English letters.

Note that you can replace two different instances of - with different characters. You can also replace each two different instances of * with different strings.

Suppose $s$ and $t$ have been transformed into $s'$ and $t'$. Now you're wondering if there's a replacement that makes $s'=t'$.

Input: The first line of input contains two integers $n$ and $m$ ($1 \leq n, m \leq 2 \cdot 10^6$) — the length of the strings $s$ and $t$, respectively.

The second line contains the string $s$ of length $n$. It is guaranteed that $s$ only consists of lowercase English letters, - and *.

The third line contains the string $t$ of length $m$. It is guaranteed that $t$ only consists of lowercase English letters, - and *.

Output: For each test case, output "Yes" if there is a replacement that makes $s'=t'$, and output "No" otherwise.

You can output "Yes" and "No" in any case (for example, strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive response).

Examples:
input:
10 10
justmonika
j-stsayori
output:
No

input:
7 8
ttk-wxx
*tt-l-xx
output:
Yes

input:
13 11
asoulwangziji
-soulg*z-y-
output:
No

input:
7 3
abc*cba
a*c
output:
No

input:
20 18
bulijiojio-dibuliduo
*li*ji-*ox*i*-du*-
output:
Yes

Note: In the second test case, we can transform both strings into ttklwxx. In $s$, - will be replaced with l. In $t$, * will be replaced by the empty string with the first and second - will be replaced with k and w respectively.

In the fifth test case, we can transform both strings into bulijiojioxdibuliduo.

Buggy Code:
import re

def replace(string, chars):
    """Replaces all occurrences of '-' in string with random lowercase English letters."""
    import random
    return re.sub('-', lambda x: random.choice(chars), string)

def main():
    # Read input
    n, m = map(int, input().split())
    s = input()
    t = input()

    # Initialize data structures
    s_list = list(s)
    t_list = list(t)
    chars = set('abcdefghijklmnopqrstuvwxyz')

    # Apply replacements to s
    i = 0
    while i < len(s_list):
        if s_list[i] == '-':
            s_list[i] = random.choice(list(chars))
        elif s_list[i] == '*':
            replacement = ''
            j = i + 1
            while j < len(s_list) and s_list[j] != '-':
                replacement += s_list[j]
                j += 1
            i += len(replacement)
            s_replacements.append(replacement)
        else:
            i += 1

    # Compare s and t after replacements
    if len(s_list) != len(t_list):
        print("No")
        return
    for i in range(len(s_list)):
        if s_list[i] != t_list[i]:
            print("No")
            return

    # All checks passed, print Yes
    print("Yes")

if __name__ == "__main__":
    main()

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
5 5
a-b*c
a-b*c
```