**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
// #define int long long
const int T = 5e5+10;
int N,A[T],Q;
struct aaa {
    int v,x;
}a[T];
// int lin[T*2],ge = 0;
// int get(int x) {
//     int ans = 0;
//     for(int i = (1<<20);i;i >>= 1)
//         if(ans+i <= ge && lin[ans+i] <= x)
//             ans += i;
//     return ans;
// }
struct node {
    int sum;
};
int B[T];
struct ttt {
    node s[T<<2];
    void build(int p,int l,int r) {
        if(l == r) {
            s[p].sum = (B[l] >= 0?0:1);
            return ;
        }
        int mid = l+r>>1;
        build(p<<1,l,mid),build(p<<1|1,mid+1,r);
        s[p].sum = s[p<<1].sum + s[p<<1|1].sum;
    }
    void change(int p,int l,int r,int x) {
        if(l == r) {
            s[p].sum = (B[l] >= 0?0:1);
            return ;
        }
        int mid = l+r>>1;
        if(x <= mid) change(p<<1,l,mid,x);
        else change(p<<1|1,mid+1,r,x);
        s[p].sum = s[p<<1].sum + s[p<<1|1].sum;
    }
    int query_l(int p,int l,int r) {
        if(s[p].sum == 0) return r+1;
        if(l == r) {
            if(s[p].sum == 0) return r+1;
            return r;
        }
        int mid = l+r>>1;
        if(s[p<<1].sum == 0) return query_l(p<<1|1,mid+1,r);
        return query_l(p<<1,l,mid);
    }
    int query_r(int p,int l,int r) {
        if(s[p].sum == 0) return l-1;
        if(l == r) {
            if(s[p].sum == 0) return l-1;
            return l;
        }
        int mid = l+r>>1;
        if(s[p<<1|1].sum == 0) return query_r(p<<1,l,mid);
        return query_r(p<<1|1,mid+1,r);
    }
}tree;
struct node2 {
    int minn,maxn;
};
struct ttt2 {
    node2 s[T<<2];
    void up(int p) {
        s[p].maxn = max(s[p<<1].maxn,s[p<<1|1].maxn);
        s[p].minn = min(s[p<<1].minn,s[p<<1|1].minn);
    }
    void build(int p,int l,int r) {
        if(l == r) {
            s[p].maxn = s[p].minn = A[l];
            return ;
        }
        int mid = l+r>>1;
        build(p<<1,l,mid),build(p<<1|1,mid+1,r);
        up(p);
    }
    void change(int p,int l,int r,int x) {
        if(l == r) {
            s[p].maxn = s[p].minn = A[x];
            return ;
        }
        int mid = l+r>>1;
        if(x <= mid) change(p<<1,l,mid,x);
        else change(p<<1|1,mid+1,r,x);
        up(p);
    }
    int query_max(int p,int l,int r,int L,int R) {
        if(L <= l && r <= R) return s[p].maxn;
        int mid = l+r>>1,maxn = -1e9;
        if(L <= mid) maxn = max(maxn,query_max(p<<1,l,mid,L,R));
        if(mid < R) maxn = max(maxn,query_max(p<<1|1,mid+1,r,L,R));
        return maxn;
    }
    int query_min(int p,int l,int r,int L,int R) {
        if(L <= l && r <= R) return s[p].minn;
        int mid = l+r>>1,minn = 1e9;
        if(L <= mid) minn = min(minn,query_min(p<<1,l,mid,L,R));
        if(mid < R) minn = min(minn,query_min(p<<1|1,mid+1,r,L,R));
        return minn;
    }
}tree2;
void solve() {
    // ge = 0;

    cin >> N;
    for(int i = 1;i <= N;i++) cin >> A[i];
    cin >> Q;
    for(int i = 1;i <= Q;i++)
        cin >> a[i].v >> a[i].x;

    for(int i = 1;i < N;i++)
        B[i] = A[i+1]-A[i];

    tree.build(1,1,N-1);
    tree2.build(1,1,N);
    int x,y;
    x = tree.query_l(1,1,N-1),y = tree.query_r(1,1,N-1);
    if(x > y) cout << -1 << " " << -1 << '\n';
    else {
        y++;
        // cout << x << " " << y+1 << '\n';
        int maxn = tree2.query_max(1,1,N,x,y);
        int minn = tree2.query_min(1,1,N,x,y);
        int ans = 0;
        for(int i = (1<<20);i;i >>= 1)
            if(ans+i < x && A[ans+i] <= minn)
                ans += i;
        int ans2 = N+1;
        for(int i = (1<<20);i;i >>= 1)
            if(ans2-i > y && A[ans2-i] >= maxn)
                ans2 -= i;
        ans++,ans2--;
        cout << ans << " " << ans2 << '\n';
    }////////////
    for(int i = 1;i <= Q;i++) {
        int v = a[i].v,x = a[i].x;
        A[v] = x;
        if(v > 1)
            B[v-1] = A[v]-A[v-1],tree.change(1,1,N-1,v-1);
        if(v < N)
            B[v] = A[v+1]-A[v],tree.change(1,1,N-1,v);
        tree2.change(1,1,N,v);

        x = tree.query_l(1,1,N-1),y = tree.query_r(1,1,N-1);
        if(x > y) cout << -1 << " " << -1 << '\n';
        else {
            // cout << x << " " << y+1 << '\n';
            y++;
            int maxn = tree2.query_max(1,1,N,x,y);
            int minn = tree2.query_min(1,1,N,x,y);
            int ans = 0;
            for(int i = (1<<20);i;i >>= 1)
                if(ans+i < x && A[ans+i] <= minn)
                    ans += i;
            int ans2 = N+1;
            for(int i = (1<<20);i;i >>= 1)
                if(ans2-i > y && A[ans2-i] >= maxn)
                    ans2 -= i;
            ans++,ans2--;
            // cout << "******** " << x << " " << y << '\n';
            // cout << "**** " << maxn << " " << minn << '\n';
            cout << ans << " " << ans2 << '\n';
        }//////////////////
    }
}
signed main() {
    ios::sync_with_stdio(false),cin.tie(0);
    int t;cin >> t;while(t--) solve();
    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:
To expose the bug in the provided buggy code, we need to craft a test input that highlights potential issues in the algorithm's logic. The main areas to focus on are:

1. **Initial Array Already Sorted**: The code should correctly handle cases where the array is already sorted.
2. **Single Element Modifications**: Changes to single elements that disrupt the sorted order.
3. **Multiple Element Modifications**: Changes to multiple elements that create complex subarray sorting requirements.
4. **Edge Cases**: Cases where the array is very small (e.g., 1 or 2 elements).

Given these considerations, let's create a small-scale test input that includes these scenarios:

```plaintext
2
5
1 2 3 4 5
3
3 10
2 6
4 1
5
1 2 2 3 4
3
2 0
4 5
3 3
```

### Explanation of the Test Input:

1. **First Test Case:**
   - **Initial Array**: `[1, 2, 3, 4, 5]` (already sorted)
   - **Modifications**:
     - **First Modification**: `3 10` (changes `a[3]` to 10) -> Array becomes `[1, 2, 10, 4, 5]`
     - **Second Modification**: `2 6` (changes `a[2]` to 6) -> Array becomes `[1, 6, 10, 4, 5]`
     - **Third Modification**: `4 1` (changes `a[4]` to 1) -> Array becomes `[1, 6, 10, 1, 5]`

2. **Second Test Case:**
   - **Initial Array**: `[1, 2, 2, 3, 4]` (almost sorted)
   - **Modifications**:
     - **First Modification**: `2 0` (changes `a[2]` to 0) -> Array becomes `[1, 0, 2, 3, 4]`
     - **Second Modification**: `4 5` (changes `a[4]` to 5) -> Array becomes `[1, 0, 2, 5, 4]`
     - **Third Modification**: `3 3` (changes `a[3]` to 3) -> Array becomes `[1, 0, 3, 5, 4]`

### Expected Output:
```plaintext
-1 -1
1 3
1 4
1 5
-1 -1
1 2
1 4
1 4
```

This test input should help identify issues in the code, especially in handling already sorted arrays and multiple modifications that disrupt the sorted order.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=500100;
int a[N];
namespace abc{
struct awa{
    int l,r,mi,mx;
    void init(int p){
        l=r=p;
        mx=mi=a[p];
    }
}z[N<<2];
awa operator+(const awa&l,const awa&r){
    return{l.l,r.r,min(l.mi,r.mi),max(l.mx,r.mx)};
}
void build(int l,int r,int rt){
    if(l==r)return z[rt].init(l);
    int mid=l+r>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    z[rt]=z[rt<<1]+z[rt<<1|1];
}
void modify(int l,int r,int rt,int p,int v){
    if(l==r){
        z[rt].mx=z[rt].mi=v;
        return;
    }
    int mid=l+r>>1;
    if(p<=mid)modify(l,mid,rt<<1,p,v);
    else modify(mid+1,r,rt<<1|1,p,v);
    z[rt]=z[rt<<1]+z[rt<<1|1];
}
int query(int l,int r,int rt,int ll,int rr){
    if(ll<=l&&r<=rr)return z[rt].mx;
    int mid=l+r>>1,ss=0;
    if(ll<=mid)ss=max(ss,query(l,mid,rt<<1,ll,rr));
    if(mid<rr)ss=max(ss,query(mid+1,r,rt<<1|1,ll,rr));
    return ss;
}
int query2(int l,int r,int rt,int ll,int rr){
    if(ll<=l&&r<=rr)return z[rt].mi;
    int mid=l+r>>1,ss=0;
    if(ll<=mid)ss=min(ss,query2(l,mid,rt<<1,ll,rr));
    if(mid<rr)ss=min(ss,query2(mid+1,r,rt<<1|1,ll,rr));
    return ss;
}
}
struct qwq{
    int l,r,sum;
    void init(int p){
        l=r=p;
        if(a[p]<=a[p+1])sum=0;
        else sum=1;
    }
}z[N<<2];
qwq operator+(const qwq&l,const qwq&r){
    return{l.l,r.r,l.sum+r.sum};
}
void build(int l,int r,int rt){
    if(l==r)return z[rt].init(l);
    int mid=l+r>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    z[rt]=z[rt<<1]+z[rt<<1|1];
}
void modify(int l,int r,int rt,int p,int v){
    if(l==r){
        z[rt].sum=v;
        return;
    }
    int mid=l+r>>1;
    if(p<=mid)modify(l,mid,rt<<1,p,v);
    else modify(mid+1,r,rt<<1|1,p,v);
    z[rt]=z[rt<<1]+z[rt<<1|1];
}
int query(int l,int r,int rt,int ll,int rr){
    if(ll<=l&&r<=rr)return z[rt].sum;
    int mid=l+r>>1,ss=0;
    if(ll<=mid)ss+=query(l,mid,rt<<1,ll,rr);
    if(mid<rr)ss+=query(mid+1,r,rt<<1|1,ll,rr);
    return ss;
}
signed main(){
    // freopen("1.in","r",stdin);
    int T;
    cin>>T;
    while(T--){
        int n,q;
        cin>>n;
        for(int i=1;i<=n;++i)cin>>a[i];
        build(1,n-1,1);
        abc::build(1,n,1);
        if(query(1,n-1,1,1,n-1)==0)cout<<"-1 -1\n";
        else{
        int l=1,r=n-1,L=1;
        while(l<=r){
            int mid=l+r>>1;
            if(query(1,n-1,1,1,mid)==0)L=mid+1,l=mid+1;
            else r=mid-1;
        }
        l=1,r=n-1;int R=n-1;
        while(l<=r){
            int mid=l+r>>1;
            if(query(1,n-1,1,mid,n-1)==0)R=mid-1,r=mid-1;
            else l=mid+1;
        }
        int MX=abc::query(1,n,1,L,R),MI=abc::query(1,n,1,L,R);
                int IL=upper_bound(a+1,a+L,MI)-a;
                if(a[1]>MI)IL=1;
                int IR=lower_bound(a+R+1,a+n+1,MX)-a-1;
                if(a[n]<MX)IR=n;
                cout<<IL<<' '<<IR<<'\n';}
        cin>>q;
        while(q--){
            int p,v;
            cin>>p>>v;
            a[p]=v;
            abc::modify(1,n,1,p,v);
            if(p!=1){
                int vx=a[p]<a[p-1];
                modify(1,n-1,1,p-1,vx);
            }if(p!=n){
                int vx=a[p]>a[p+1];
                modify(1,n-1,1,p,vx);
            }
            if(query(1,n-1,1,1,n-1)==0)cout<<"-1 -1\n";
            else{
                int l=1,r=n-1,L=1;
                while(l<=r){
                    int mid=l+r>>1;
                    if(query(1,n-1,1,1,mid)==0)L=mid+1,l=mid+1;
                    else r=mid-1;
                }
                l=1,r=n-1;int R=n-1;
                while(l<=r){
                    int mid=l+r>>1;
                    if(query(1,n-1,1,mid,n-1)==0)R=mid-1,r=mid-1;
                    else l=mid+1;
                }
                ++R;
                int MX=abc::query(1,n,1,L,R),MI=abc::query(1,n,1,L,R);
                int IL=upper_bound(a+1,a+L,MI)-a;
                if(a[1]>MI)IL=1;
                int IR=lower_bound(a+R+1,a+n+1,MX)-a-1;
                if(a[n]<MX)IR=n;
                cout<<IL<<' '<<IR<<'\n';
            }
        }
        // cerr<<'\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
2
5
1 2 3 4 5
3
2 1
4 1
1 1
5
5 4 3 2 1
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=500010;
int T,n,q;
int a[N],d[N];
int trmx[N*4],trmn[N*4],dmn[N*4];
void maintain(int o){
    trmx[o]=max(trmx[o<<1],trmx[o<<1|1]);
    trmn[o]=min(trmn[o<<1],trmn[o<<1|1]);
    dmn[o]=min(dmn[o<<1],dmn[o<<1|1]);
    return;
}
void build(int o,int l,int r){
    if(l==r){
        trmx[o]=trmn[o]=a[l];
        dmn[o]=d[l];
        return;
    }
    int mid=(l+r)>>1;
    build(o<<1,l,mid);
    build(o<<1|1,mid+1,r);
    maintain(o);
    return;
}
void updatea(int o,int l,int r,int x){
    if(l==r){
        trmn[o]=trmx[o]=a[x];
        return;
    }
    int mid=(l+r)>>1;
    if(x<=mid)
        updatea(o<<1,l,mid,x);
    if(x>mid)
        updatea(o<<1|1,mid+1,r,x);
    maintain(o);
    return;
}
void updated(int o,int l,int r,int x){
    if(l==r){
        dmn[o]=d[x];
        return;
    }
    int mid=(l+r)>>1;
    if(x<=mid)
        updated(o<<1,l,mid,x);
    if(x>mid)
        updated(o<<1|1,mid+1,r,x);
    maintain(o);
    return;
}
int query1l(int o,int l,int r){
    if(l==r){
        if(d[l]>=0)
            return l;
        return l-1;
    }
    int mid=(l+r)>>1;
    if(dmn[o<<1]>=0)
        return query1l(o<<1|1,mid+1,r);
    return query1l(o<<1,l,mid);
}
int query1r(int o,int l,int r){
    if(l==r){
        if(d[l]>=0)
            return l-1;
        return l;
    }
    int mid=(l+r)>>1;
    if(dmn[o<<1|1]>=0)
        return query1r(o<<1,l,mid);
    return query1r(o<<1,mid+1,r);
}
int query2l(int o,int l,int r,int x,int lmx,int rmn){
    if(l==r){
        if(max(lmx,a[l])<=rmn)
            return l+1;
        return l;
    }
    int mid=(l+r)>>1;
    if(x<=mid||max(lmx,trmx[o<<1])>min(rmn,trmn[o<<1|1]))
        return query2l(o<<1,l,mid,x,lmx,min(rmn,trmn[o<<1|1]));
    return query2l(o<<1|1,mid+1,r,x,max(lmx,trmx[o<<1]),rmn);
}
int query2r(int o,int l,int r,int x,int lmx,int rmn){
    if(l==r){
        if(lmx<=min(rmn,a[l]))
            return l-1;
        return l;
    }
    int mid=(l+r)>>1;
    if(x>mid||max(lmx,trmx[o<<1])>min(rmn,trmn[o<<1|1]))
        return query2r(o<<1|1,mid+1,r,x,max(lmx,trmx[o<<1]),rmn);
    return query2r(o<<1,l,mid,x,lmx,min(rmn,trmn[o<<1|1]));
}
void Getans(){
    if(dmn[1]>=0){
        printf("-1 -1\n");
        return;
    }
    int l=query1l(1,1,n),r=query1r(1,1,n);
    if(query2r(1,1,n,r,-0x7f7f7f7f,0x7f7f7f7f)<r){
        printf("-2 -2\n");
    }
    printf("%d %d\n",query2l(1,1,n,l,-0x7f7f7f7f,0x7f7f7f7f),query2r(1,1,n,r,-0x7f7f7f7f,0x7f7f7f7f));
    return;
}
int main(){
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        a[0]=-1e9;
        for(int i=1;i<=n;i++)
            d[i]=a[i]-a[i-1];
        build(1,1,n);
        scanf("%d",&q);
        Getans();
        while(q--){
            int p,v;
            scanf("%d%d",&p,&v);
            a[p]=v;
            d[p]=a[p]-a[p-1];
            d[p+1]=a[p+1]-a[p];
            updatea(1,1,n,p);
            updated(1,1,n,p);
            if(p+1<=n)
                updated(1,1,n,p+1);
            Getans();
        }
    }
    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
2
5
1 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
print("""-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-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
2
5
1 2 3 4 5
3
3 6
2 6
1 6
4
5 4 3 2
2
1 1
4 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import bisect
import sys, os, io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

def f(u, v):
    return u << 20 ^ v

def update(i, x, y):
    i ^= l1
    mi[i], ma[i] = x, y
    i >>= 1
    while i:
        mi[i] = min(mi[i << 1], mi[i << 1 ^ 1])
        ma[i] = max(ma[i << 1], ma[i << 1 ^ 1])
        i >>= 1
    return

def get_sum(i):
    s = 0
    while i > 0:
        s += tree[i]
        i -= i & -i
    return s

def add(i, x):
    while i < len(tree):
        tree[i] += x
        i += i & -i

t = int(input())
ans = []
inf = pow(10, 9) + 1
for _ in range(t):
    n = int(input())
    a = [-inf] + list(map(int, input().split())) + [inf]
    q = int(input())
    q0 = [0] * (q << 1)
    x = [0] * (n + q + 1)
    for i in range(n + 1):
        x[i] = f(a[i] + inf, i)
    for i in range(q):
        pos, val = map(int, input().split())
        x[i + n] = f(val + inf, pos)
        q0[i << 1], q0[i << 1 ^ 1] = pos, val
    l1 = pow(2, (n + q + 1).bit_length())
    l2 = 2 * l1
    mi, ma = [inf] * l2, [-inf] * l2
    tree = [0] * (n + q + 5)
    x.sort()
    for i in range(1, n + 1):
        a[i] = bisect.bisect_left(x, f(a[i] + inf, i))
        tree[a[i]] += 1
    for i in range(1, len(tree)):
        j = i + (i & -i)
        if j < len(tree):
            tree[j] += tree[i]
    for i in range(2, n):
        if a[i - 1] > a[i] < a[i + 1] or a[i - 1] < a[i] > a[i + 1]:
            update(a[i], a[i], a[i])
    ans0 = ["-1 -1" for _ in range(q + 1)]
    if mi[1] ^ inf:
        l, r = get_sum(mi[1]), get_sum(ma[1])
        ans0[0] = " ".join(map(str, (l, r)))
    for i in range(q):
        pos, val = q0[i << 1], q0[i << 1 ^ 1]
        for j in range(max(pos - 1, 1), min(pos + 2, n + 1)):
            if a[j - 1] > a[j] < a[j + 1] or a[j - 1] < a[j] > a[j + 1]:
                update(a[j], inf, -inf)
        add(a[pos], -1)
        a[pos] = bisect.bisect_left(x, f(val + inf, pos))
        for j in range(max(pos - 1, 1), min(pos + 2, n + 1)):
            if a[j - 1] > a[j] < a[j + 1] or a[j - 1] < a[j] > a[j + 1]:
                update(a[j], a[j], a[j])
        add(a[pos], 1)
        if mi[1] ^ inf:
            l, r = get_sum(mi[1]), get_sum(ma[1])
            ans0[i + 1] = " ".join(map(str, (l, r)))
    ans.append("\n".join(ans0))
sys.stdout.write("\n".join(ans))

**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
2
5
1 2 3 4 5
3
1 6
2 7
3 8
5
5 4 3 2 1
3
1 1
2 2
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import java.util.*;

public class sortingProblemAgain {

    public static void ans(int max, int min, int arr[], int n){
                // System.out.println("max:"+max+" min:"+min+" n:"+n);
                int l = Integer.MAX_VALUE;
                int r = Integer.MIN_VALUE;

                int sorted = 0;

                if(arr[0]!=min && arr[n-1]!=max){

                    l = 0;
                    r = n-1;

                }else if(arr[0]!=min && arr[n-1]==max){

                    l = 0;

                    int maxarr[] = new int[n];
                    maxarr[0] = arr[0];

                    for(int i=1; i<n; i++){
                        maxarr[i] = Math.max(arr[i], maxarr[i-1]);
                    }

                    for(int j=n-1; j>0; j--){
                        if(arr[j]!=maxarr[j]){
                            r = j;
                            sorted = -1;
                            break;
                        }else{
                            sorted = 1;
                        }
                    }

                }else if(arr[0]==min && arr[n-1]!=max){

                    r = n-1;

                    int minRev[] = new int[n];
                    minRev[n-1] = arr[n-1];

                    for(int i=n-2; i>=0; i--){
                        minRev[i] = Math.min(arr[i], minRev[i+1]);
                    }

                    for(int j=1; j<n-1; j++){
                        if(arr[j]!=minRev[j]){
                            l = j;
                            sorted = -1;
                            break;
                        }else{
                            sorted = 1;
                        }
                    }

                }else if(arr[0]==min && arr[n-1]==max){

                    int minRev[] = new int[n];
                    minRev[n-1] = arr[n-1];

                    for(int i=n-2; i>=0; i--){
                        minRev[i] = Math.min(arr[i], minRev[i+1]);
                    }

                    for(int j=1; j<n-1; j++){
                        if(arr[j]!=minRev[j]){
                            l = j;
                            sorted = -1;
                            break;
                        }else{
                            sorted = 1;
                        }
                    }

                    int maxarr[] = new int[n];
                    maxarr[0] = arr[0];

                    for(int i=1; i<n; i++){
                        maxarr[i] = Math.max(arr[i], maxarr[i-1]);
                    }

                    for(int j=n-1; j>0; j--){
                        if(arr[j]!=maxarr[j]){
                            r = j;
                            sorted = -1;
                            break;
                        }else{
                            sorted = 1;
                        }
                    }

                }

                if(sorted==1){
                    l = -1;
                    r = -1;
                    System.out.println(l+" "+r);
                }else{
                    System.out.println((l+1)+" "+(r+1));
                }

    }

    public static void main(String args[]){

        try(Scanner sc = new Scanner(System.in)){

            int test = sc.nextInt();

            while((test--)>0){

                int n = sc.nextInt();

                int arr[] = new int[n];

                int min = Integer.MAX_VALUE;
                int max = Integer.MIN_VALUE;

                for(int i=0; i<n; i++){

                    arr[i] = sc.nextInt();

                    min = Math.min(arr[i], min);
                    max = Math.max(arr[i], max);

                }

                ans(max, min, arr, n);

                int q = sc.nextInt();

                while((q--)>0){

                    int pos = sc.nextInt();
                    int change = sc.nextInt();

                    arr[pos-1] = change;

                    max = Math.max(max, change);
                    min = Math.min(min, change);

                    ans(max, min, arr, 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
2
5
1 2 3 4 5
3
1 5
2 4
3 3
5
5 4 3 2 1
3
1 1
2 2
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import java.io.*;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main5 {

    static TreeMap<Integer, Integer> map, map2;
    static TreeSet<Integer> set;
    static int[] a;
    static int n;

    public static void del(int x) {
        if (x > 1 && a[x-1] > a[x]) {
            set.remove(x-1);
            map.put(a[x-1], map.get(a[x-1]) - 1);
            map2.put(a[x], map2.get(a[x]) - 1);
            if (map.get(a[x-1]) == 0) map.remove(a[x-1]);
            if (map2.get(a[x]) == 0) map2.remove(a[x]);
        }
        if (x < n && a[x] > a[x+1]) {
            set.remove(x);
            map.put(a[x], map.get(a[x]) - 1);
            map2.put(a[x+1], map2.get(a[x+1]) - 1);
            if (map.get(a[x]) == 0) map.remove(a[x]);
            if (map2.get(a[x+1]) == 0) map2.remove(a[x+1]);
        }
    }

    public static void add(int x) {
        if (x > 1 && a[x-1] > a[x]) {
            set.add(x-1);
            map.put(a[x-1], map.getOrDefault(a[x-1], 0) + 1);
            map2.put(a[x], map2.getOrDefault(a[x], 0) + 1);
        }
        if (x < n && a[x] > a[x+1]) {
            set.add(x);
            map.put(a[x], map.getOrDefault(a[x], 0) + 1);
            map2.put(a[x+1], map2.getOrDefault(a[x+1], 0) + 1);
        }
    }

    public static int[] get() {
        if (set.size() == 0) {
            return new int[]{-1, -1};
        }
        int[] ans = new int[2];
        int x1 = set.first(), x2 = set.last() + 1;
        int l = 1, r = x1;
        while (l < r) {
            int mid = l + r >> 1;
            if (a[mid] > a[x1]) r = mid;
            else l = mid + 1;
        }
        ans[0] = r;
        l = x2;
        r = n;
        while (l <= r) {
            int mid = l + r >> 1;
            if (a[mid] >= a[x2-1]) {
                r = mid - 1;
            }else {
                ans[1] = mid;
                l = mid + 1;
            }
        }
        return ans;
    }

    public static void solve() throws IOException{
        n = in.nextInt();
        a = new int[n+1];
        for (int i = 1; i <= n; i++) {
            a[i] = in.nextInt();
        }
        map = new TreeMap<>();
        map2 = new TreeMap<>();
        set = new TreeSet<>();
        for (int i = 1; i < n; i++) {
            if (a[i] > a[i+1]) {
                set.add(i);
                map.put(a[i], map.getOrDefault(a[i], 0) + 1);
                map2.put(a[i+1], map2.getOrDefault(a[i+1], 0) + 1);
            }
        }
        int[] t = get();
        out.println(t[0] + " " + t[1]);
        int q = in.nextInt();
        while (q-- > 0) {
            int x = in.nextInt();
            int v = in.nextInt();
            del(x);
            a[x] = v;
            add(x);
            int[] ans = get();
            out.println(ans[0] + " " + ans[1]);
        }
    }

    static boolean MULTI_CASE = true;

    public static void main(String[] args) throws IOException {
        if (MULTI_CASE) {
            int T = in.nextInt();
            for (int i = 0; i < T; ++i) {
                solve();
            }
        } else {
            solve();
        }
        out.close();
    }

    static InputReader in = new InputReader();
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    static class InputReader {
        private StringTokenizer st;
        private BufferedReader bf;

        public InputReader() {
            bf = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(bf.readLine());
            }
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return bf.readLine();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

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

/*
 */

**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
2
5
1 2 3 4 5
1
3 0
5
1 1 1 1 1
1
3 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
def find_min_subarray(arr):
    n = len(arr)

    if all(arr[i] <= arr[i+1] for i in range(n-1)):
        return (-1, -1)

    # Find the first position where the array is out of order from the left
    l = 0
    while l < n - 1 and arr[l] <= arr[l + 1]:
        l += 1

    # Find the first position where the array is out of order from the right
    r = n - 1
    while r > 0 and arr[r] >= arr[r - 1]:
        r -= 1

    # Expand the window to include any additional elements needed to maintain order
    while l > 0 and arr[l - 1] > arr[r]:
        l -= 1
    while r < n - 1 and arr[r + 1] < arr[l]:
        r += 1

    return (l + 1, r + 1)  # Convert to 1-based index

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    idx = 0
    t = int(data[idx])
    idx += 1
    results = []

    for _ in range(t):
        n = int(data[idx])
        idx += 1
        a = list(map(int, data[idx:idx + n]))
        idx += n
        q = int(data[idx])
        idx += 1

        queries = []
        for __ in range(q):
            pos = int(data[idx]) - 1
            val = int(data[idx + 1])
            queries.append((pos, val))
            idx += 2

        # Initial state
        res = find_min_subarray(a)
        results.append(f"{res[0]} {res[1]}")

        # Handle each query
        for pos, val in queries:
            a[pos] = val
            res = find_min_subarray(a)
            results.append(f"{res[0]} {res[1]}")

    print("\n".join(results))

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
1
5
1 2 3 4 5
5
1 10
2 9
3 8
4 7
5 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN = 5e5 + 5;

int n, q;
long long a[MAXN], sorted[MAXN];

pair<int, int> find_unsorted_range() {
    int left = n + 1, right = 0;
    for (int i = 1; i <= n; i++) {
        if (a[i] != sorted[i]) {
            left = min(left, i);
            break;
        }
    }
    for (int i = n; i >= 1; i--) {
        if (a[i] != sorted[i]) {
            right = max(right, i);
            break;
        }
    }
    if (left > right) return {-1, -1};
    return {left, right};
}

void solve() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        sorted[i] = a[i];
    }
    sort(sorted + 1, sorted + n + 1);

    auto [l, r] = find_unsorted_range();
    cout << l << " " << r << "\n";

    cin >> q;
    while (q--) {
        int pos;
        long long val;
        cin >> pos >> val;

        // Update the arrays
        a[pos] = val;
        for (int i = 1; i <= n; i++) {
            sorted[i] = a[i];
        }
        sort(sorted + 1, sorted + n + 1);

        auto [l, r] = find_unsorted_range();
        cout << l << " " << r << "\n";
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;
    while (t--) solve();

    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
1
5
1 2 3 4 5
5
1 5
2 4
3 3
4 2
5 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
def find_unsorted_subarray(arr):
    n = len(arr)
    if n <= 1:
        return (-1, -1)

    # Find the first element which is out of order from the beginning
    for l in range(n - 1):
        if arr[l] > arr[l + 1]:
            break
    else:
        return (-1, -1)

    # Find the first element which is out of order from the end
    for r in range(n - 1, 0, -1):
        if arr[r] < arr[r - 1]:
            break

    # Expand the window to include all necessary elements
    min_val = min(arr[l:r+1])
    max_val = max(arr[l:r+1])

    while l > 0 and arr[l - 1] > min_val:
        l -= 1
    while r < n - 1 and arr[r + 1] < max_val:
        r += 1

    return (l + 1, r + 1)

def process_queries(arr, queries):
    results = []
    for pos, val in queries:
        arr[pos - 1] = val
        result = find_unsorted_subarray(arr)
        results.append(result)
    return results

# Example usage
if __name__ == "__main__":
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        index += 1
        arr = list(map(int, data[index:index + n]))
        index += n
        q = int(data[index])
        index += 1
        queries = []

        for _ in range(q):
            pos = int(data[index])
            val = int(data[index + 1])
            index += 2
            queries.append((pos, val))

        initial_result = find_unsorted_subarray(arr)
        results.append(initial_result)
        results.extend(process_queries(arr, queries))

    for res in results:
        print(*res)

**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
2
5
1 2 3 4 5
3
2 1
4 1
1 1
4
5 4 3 2
2
1 1
4 6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
def find_minimal_subarray_to_sort(arr):
    n = len(arr)

    # Step 1: Find the first place where arr[i] > arr[i+1]
    l = -1
    for i in range(n - 1):
        if arr[i] > arr[i + 1]:
            l = i
            break

    # If no such index is found, the array is already sorted
    if l == -1:
        return -1, -1

    # Step 2: Find the last place where arr[j] < arr[j-1]
    r = -1
    for j in range(n - 1, 0, -1):
        if arr[j] < arr[j - 1]:
            r = j
            break

    # Step 3: Find the min and max of the subarray [l, r]
    subarray_min = min(arr[l:r + 1])
    subarray_max = max(arr[l:r + 1])

    # Step 4: Extend l to the left while arr[l] > subarray_min
    while l > 0 and arr[l - 1] > subarray_min:
        l -= 1

    # Step 5: Extend r to the right while arr[r] < subarray_max
    while r < n - 1 and arr[r + 1] < subarray_max:
        r += 1

    return l + 1, r + 1  # Convert to 1-based indexing

def solve():
    import sys
    input = sys.stdin.read().split()
    ptr = 0

    t = int(input[ptr])
    ptr += 1

    results = []

    for _ in range(t):
        n = int(input[ptr])
        ptr += 1

        arr = list(map(int, input[ptr:ptr + n]))
        ptr += n

        q = int(input[ptr])
        ptr += 1

        # Initial check
        l, r = find_minimal_subarray_to_sort(arr)
        results.append((l, r))

        for _ in range(q):
            pos = int(input[ptr]) - 1  # Convert to 0-based indexing
            val = int(input[ptr + 1])
            ptr += 2

            # Update the array
            arr[pos] = val

            # Find the new minimal subarray to sort
            l, r = find_minimal_subarray_to_sort(arr)
            results.append((l, r))

    for l, r in results:
        print(l, r)

if __name__ == "__main__":
    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:
```plaintext
2
4
1 2 3 4
3
1 5
2 6
3 7
5
1 2 3 4 5
4
1 6
2 7
3 8
4 9
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import java.util.*;
public class sortingproblem {
    public static void func(int arr[]){
        int brr[]=arr.clone();
        // brr=arr;
        Arrays.sort(brr);
        // for(int i=0;i<brr.length;i++){
        //     System.out.print(brr[i]+" ");
        // }
        // System.out.println();
        // for(int i=0;i<brr.length;i++){
        //     System.out.print(arr[i]+" ");
        // }
        int i=0;
        int j=arr.length-1;
        int l=-1;
        int r=-1;
        // System.out.println(brr);
        // while(i<=j){
        //     if(arr[i]!=brr[i] && l==-1){
        //         l=i+1;
        //     }
        //     if(arr[j]!=brr[j] && r==-1){
        //         r=j+1;
        //     }
        //     i++;
        //     j--;
        // }
        while(i<arr.length){
            if(arr[i]!=brr[i]){
                l=i+1;
                break;
            }
            i++;
        }
        while(j>=0){
            if(arr[j]!=brr[j]){
                r=j+1;
                break;
            }
            j--;
        }
        if(l==-1 && r==-1){
            System.out.println(l+" "+r);
        }
        else if(l==-1){
            System.out.println(0+" "+r);
        }
        else if(r==-1){
            System.out.println(l+" "+arr.length);
        }
        else{
            System.out.println(l+" "+r);
        }
    }
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        // int arr[]={2,2,3,4,5};
        int testcase=sc.nextInt();
        while(testcase>0){
            int n=sc.nextInt();
            int arr[]=new int[n];
            for(int i=0;i<n;i++){
                arr[i]=sc.nextInt();
            }
            func(arr);
            int t=sc.nextInt();
            while(t>0){
                int pos=sc.nextInt();
                int val=sc.nextInt();
                arr[pos-1]=val;
                func(arr);
                t--;
            }
            testcase--;
        }

    }

}

**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
1
5
1 2 3 4 5
3
1 5
2 4
3 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

// Function to find the smallest subarray that, if sorted, will make the whole array sorted
pair<int, int> findUnsortedSubarray(const vector<int>& a) {
    int n = a.size();
    int left = -1, right = -1;

    // Find the first position from the left where order breaks
    for (int i = 0; i < n - 1; ++i) {
        if (a[i] > a[i + 1]) {
            left = i;
            break;
        }
    }

    // If no break is found, the array is already sorted
    if (left == -1) {
        return {-1, -1};
    }

    // Find the first position from the right where order breaks
    for (int i = n - 1; i > 0; --i) {
        if (a[i] < a[i - 1]) {
            right = i;
            break;
        }
    }

    // Find the min and max in the subarray a[left:right]
    int subarray_min = *min_element(a.begin() + left, a.begin() + right + 1);
    int subarray_max = *max_element(a.begin() + left, a.begin() + right + 1);

    // Extend the left boundary if needed
    while (left > 0 && a[left - 1] > subarray_min) {
        left--;
    }

    // Extend the right boundary if needed
    while (right < n - 1 && a[right + 1] < subarray_max) {
        right++;
    }

    return {left + 1, right + 1}; // Convert to 1-based indexing
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;

        vector<int> a(n);
        for (int i = 0; i < n; ++i) {
            cin >> a[i];
        }

        // Initial result
        auto result = findUnsortedSubarray(a);
        cout << result.first << " " << result.second << endl;

        int q;
        cin >> q;
        while (q--) {
            int pos, val;
            cin >> pos >> val;
            --pos; // Convert to 0-based indexing

            a[pos] = val; // Apply modification
            result = findUnsortedSubarray(a);
            cout << result.first << " " << result.second << 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
2
5
5 4 3 2 1
2
3 5
1 3
4
4 3 2 1
3
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import java.util.*;

public class SortingProblemAgain {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while (t-- > 0) {
            int n = sc.nextInt();
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
            }

            int q = sc.nextInt();
            List<int[]> modifications = new ArrayList<>();
            for (int i = 0; i < q; i++) {
                int pos = sc.nextInt() - 1;
                int val = sc.nextInt();
                modifications.add(new int[]{pos, val});
            }

            List<int[]> results = new ArrayList<>();
            results.add(findMinimumSubarray(a));

            for (int[] mod : modifications) {
                int pos = mod[0];
                int val = mod[1];
                a[pos] = val;
                results.add(findMinimumSubarray(a));
            }

            for (int[] res : results) {
                System.out.println(res[0] + " " + res[1]);
            }
        }
    }

    private static int[] findMinimumSubarray(int[] a) {
        int n = a.length;
        int l = -1, r = -1;

        for (int i = 0; i < n - 1; i++) {
            if (a[i] > a[i + 1]) {
                l = i;
                break;
            }
        }

        if (l == -1) return new int[]{-1, -1};

        for (int i = n - 1; i > 0; i--) {
            if (a[i] < a[i - 1]) {
                r = i;
                break;
            }
        }

        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        for (int i = l; i <= r; i++) {
            min = Math.min(min, a[i]);
            max = Math.max(max, a[i]);
        }

        for (int i = 0; i < l; i++) {
            if (a[i] > min) {
                l = i;
                break;
            }
        }

        for (int i = n - 1; i >= r + 1; i--) {
            if (a[i] < max) {
                r = i;
                break;
            }
        }

        return new int[]{l + 1, r + 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
2
5
1 2 3 4 5
3
2 6
3 7
1 8
5
5 4 3 2 1
3
1 2
2 3
3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
def find_min_unsorted_subarray(a):
    n = len(a)
    l, r = -1, -1

    # Find the first position where the order is violated
    for i in range(n - 1):
        if a[i] > a[i + 1]:
            l = i
            break

    # If no such position is found, the array is already sorted
    if l == -1:
        return -1, -1

    # Find the last position where the order is violated
    for j in range(n - 1, 0, -1):
        if a[j] < a[j - 1]:
            r = j
            break

    # Now expand the l and r to include any out of order elements within [l, r]
    min_val = min(a[l:r+1])
    max_val = max(a[l:r+1])

    while l > 0 and a[l-1] > min_val:
        l -= 1

    while r < n - 1 and a[r+1] < max_val:
        r += 1

    return l, r

def process_test_case(n, a, q, modifications):
    results = []

    # Initially, check the array and find the minimum unsorted subarray
    l, r = find_min_unsorted_subarray(a)
    results.append((l + 1, r + 1) if l != -1 else (-1, -1))

    for pos, val in modifications:
        # Apply the modification
        a[pos - 1] = val

        # Check again for the minimum unsorted subarray
        l, r = find_min_unsorted_subarray(a)
        results.append((l + 1, r + 1) if l != -1 else (-1, -1))

    return results

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        index += 1
        a = list(map(int, data[index:index + n]))
        index += n
        q = int(data[index])
        index += 1
        modifications = []
        for __ in range(q):
            pos = int(data[index])
            val = int(data[index + 1])
            index += 2
            modifications.append((pos, val))

        results.extend(process_test_case(n, a, q, modifications))

    for result in results:
        print(result[0], result[1])

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
2
5
1 2 3 4 5
3
3 0
3 6
3 3
6
5 4 3 2 1 0
5
1 0
2 1
3 2
4 3
5 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Sorting Problem Again
time_limit_ms: 2500
memory_limit_mb: 256
Description: You have an array $a$ of $n$ elements. There are also $q$ modifications of the array. Before the first modification and after each modification, you would like to know the following:

What is the minimum length subarray that needs to be sorted in non-decreasing order in order for the array $a$ to be completely sorted in non-decreasing order?

More formally, you want to select a subarray of the array $(l, r)$ with the minimum value of $r - l + 1$. After that, you will sort the elements $a_{l}, a_{l + 1}, \ldots, a_{r}$ and want the condition $a_{i} \le a_{i + 1}$ to hold for all $1 \le i < n$. If the array is already sorted in non-decreasing order, then $l$ and $r$ should be considered as equal to $-1$.

Note that finding such $(l, r)$ does not change the array in any way. The modifications themselves take the form: assign $a_{pos} = x$ for given $pos$ and $x$.

Input: Each test consists of several test cases. The first line contains an integer $t$ ($1 \le t \le 10$) — the number of test cases. Then follows the description of test cases.

The first line of each test case contains a single integer $n$ ($1 \le n \le 5 \cdot 10^{5}$).

The second line of each test case contains $n$ integers $a_{i}$ ($0 \le |a_{i}| \le 10^{9}$) — the initial elements of the array $a$.

The third line of each test case contains a number $q$ ($0 \le q \le 5 \cdot 10^{5}$) — the number of modifications to the array.

The following $q$ lines of each test case contain two integers $pos_{i}$ ($1 \le pos_{i} \le n$) and $val_{i}$ ($0 \le |val_{i}| \le 10^{9}$) — this means that for the $i$-th modification, $a_{pos_{i}}$ is assigned the value $val_{i}$.

It is guaranteed that the sum of $n$ and the sum of $q$ for all test cases does not exceed $5 \cdot 10^{5}$.

Output: For each test case, output $q + 1$ lines. Each line should contain $2$ integers $l, r$ — the boundaries of the minimum subarray, such that sorting it will make the array $a$ completely sorted. If $a$ is already sorted, then output $l = -1$, $r = -1$.

Examples:
input:
2
5
2 2 3 4 5
3
2 1
4 1
1 1
5
1 2 3 4 5
9
1 4
2 3
5 2
3 1
1 1
5 1
4 1
3 1
2 1
output:
-1 -1
1 2
1 4
3 4
-1 -1
1 3
1 3
1 5
1 5
2 5
2 5
2 5
2 5
-1 -1

Note: Let's consider the first test case:

  * Initially, the array is sorted in non-decreasing order: $[2, 2, 3, 4, 5]$    * After the first query, the array looks like this: $[\color{red}{2}, \color{red}{1}, 3, 4, 5]$.    * After the second query, the array looks like this: $[\color{red}{2}, \color{red}{1}, \color{red}{3}, \color{red}{1}, 5]$.    * After the third query, the array looks like this: $[1, 1, \color{red}{3}, \color{red}{1}, 5]$.

The red segments indicate the subarrays that need to be sorted in order for the entire array to be sorted in non-decreasing order.

Buggy Code:
import bisect
import heapq
import math
import sys
from collections import deque
from functools import cache, lru_cache
from random import seed, randint

if 'MY_LOCAL' in sys.argv:
    from my.utils import *

input = lambda: sys.stdin.readline()
in_str = lambda: input().strip()
in_int = lambda: int(input())
in_ints = lambda: list(map(int, input().split()))
in_intt = lambda: tuple(map(int, input().split()))
print_bool = lambda v: print('YES' if v else 'NO')
print_list = lambda v: print(' '.join(map(str, v)))

infp = 10 ** 10
infn = -infp

def solve():
    n = in_int()
    aa = in_ints()
    pv = [in_intt() for _ in range(in_int())]

    def find():
        mx = infn
        r = -1
        for i in range(n):
            if aa[i] < mx:
                r = i
            mx = max(aa[i], mx)
        if r == -1:
            print(-1, -1)
            return

        mn = infp
        l = -1
        for i in range(n - 1, -1, -1):
            if aa[i] > mn:
                l = i
            mn = min(aa[i], mn)
        print(l + 1, r + 1)

    find()
    for p, v in pv:
        aa[p - 1] = v
        find()

for _ in range(in_int()):
    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 create a challenging test input that exposes potential bugs in the provided buggy code, we need to consider edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Initial Sorted Array**: Start with an array that is already sorted. This will test the initial condition where the output should be `-1 -1`.

2. **Single Element Array**: Test with a single element array. This should also output `-1 -1` since a single element is trivially sorted.

3. **Array with Duplicates**: Test with an array containing duplicate elements. This will check how the code handles cases where the same value appears multiple times.

4. **Array with a Single Out-of-Order Element**: Test with an array that has one element out of order. This will check if the code can correctly identify the minimal subarray that needs to be sorted.

5. **Array with Multiple Out-of-Order Elements**: Test with an array that has multiple elements out of order. This will check if the code can handle more complex scenarios.

6. ** modifications**: Include modifications that change the array in such a way that the minimal subarray changes over time.

Based on these considerations, here is a small-scale test input:

```plaintext
3
5
1 2 3 4 5
3
3 6
1 6
5 6
1
1
0
3
1 1 1
1
2 2
4
1 2 3 4
1
2 5
4
1 2 3 4
1
2 0
```

### Explanation:

1. **First Test Case**:
   - Initial array: `[1, 2, 3, 4, 5]` (sorted)
   - Modifications:
     - `3 6` (array becomes `[1, 2, 6, 4, 5]`)
     - `1 6` (array becomes `[6, 2, 6, 4, 5]`)
     - `5 6` (array becomes `[6, 2, 6, 4, 6]`)

2. **Second Test Case**:
   - Initial array: `[1]` (single element)
   - No modifications.

3. **Third Test Case**:
   - Initial array: `[1, 1, 1]` (all elements are the same)
   - Modification:
     - `2 2` (array becomes `[1, 2, 1]`)

4. **Fourth Test Case**:
   - Initial array: `[1, 2, 3, 4]` (sorted)
   - Modification:
     - `2 5` (array becomes `[1, 5, 3, 4]`)

5. **Fifth Test Case**:
   - Initial array: `[1, 2, 3, 4]` (sorted)
   - Modification:
     - `2 0` (array becomes `[1, 0, 3, 4]`)

This test input covers a variety of scenarios, including sorted arrays, single-element arrays, arrays with duplicates, and arrays with out-of-order elements. It also includes modifications that change the array in different ways to test the robustness of the code.