**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Funny Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Vanya has a graph with $$$n$$$ vertices (numbered from $$$1$$$ to $$$n$$$) and an array $$$a$$$ of $$$n$$$ integers; initially, there are no edges in the graph. Vanya got bored, and to have fun, he decided to perform $$$n - 1$$$ operations.

Operation number $$$x$$$ (operations are numbered in order starting from $$$1$$$) is as follows:

- Choose $$$2$$$ different numbers $$$1 \leq u,v \leq n$$$, such that $$$|a_u - a_v|$$$ is divisible by $$$x$$$.
- Add an undirected edge between vertices $$$u$$$ and $$$v$$$ to the graph.

Help Vanya get a connected$$$^{\text{∗}}$$$ graph using the $$$n - 1$$$ operations, or determine that it is impossible.

Input: Each test consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^{3}$$$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains the number $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of vertices in the graph.

The second line of each test case contains $$$n$$$ numbers $$$a_1, a_2, \cdots a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$.

Output: For each test case, if there is no solution, then output "No" (without quotes).

Otherwise, output "Yes" (without quotes), and then output $$$n - 1$$$ lines, where in the $$$i$$$-th line, output the numbers $$$u$$$ and $$$v$$$ that need to be chosen for operation $$$i$$$.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
2
1 4
4
99 7 1 13
5
10 2 31 44 73
5
87 6 81 44 32
5
62 35 33 79 16
5
6 51 31 69 42
5
52 63 25 21 5
12
33 40 3 11 31 43 37 8 50 5 12 22
output:
YES
2 1
YES
4 1
2 1
3 2
YES
5 1
4 1
3 1
2 1
YES
4 1
3 1
2 1
5 4
YES
3 1
5 1
2 1
4 2
YES
4 1
5 1
2 1
3 2
YES
2 1
5 2
3 1
4 3
YES
9 1
12 9
11 1
10 1
6 1
7 6
2 1
8 2
5 2
3 1
4 1

Note: Let's consider the second test case.

- First operation $$$(x = 1)$$$: we can connect vertices $$$4$$$ and $$$1$$$, since $$$|a_4 - a_1| = |13 - 99| = |-86| = 86$$$, and $$$86$$$ is divisible by $$$1$$$.

- Second operation $$$(x = 2)$$$: we can connect vertices $$$2$$$ and $$$1$$$, since $$$|a_2 - a_1| = |7 - 99| = |-92| = 92$$$, and $$$92$$$ is divisible by $$$2$$$.

- Third operation $$$(x = 3)$$$: we can connect vertices $$$3$$$ and $$$2$$$, since $$$|a_3 - a_2| = |1 - 7| = |-6| = 6$$$, and $$$6$$$ is divisible by $$$3$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;

// #define int long long
#define fi first
#define se second
#define endl '\n'
#define lc (u << 1)
#define rc (u << 1 | 1)
#define pb push_back
#define eb emplace_back
#define ep emplace
using LL = long long;
using ull = unsigned long long;
using pii = pair<int, int>;

/******************************************************************/
#define debug(x...)                  \
    do {                           \
        std::cout << #x << " -> "; \
        err(x);                    \
    } while (0)

void err() {
    std::cout << '\n';
}

template <class T, class... Ts>
void err(T arg, Ts ...args) {
    std::cout << arg << ' ';
    err(args...);
}

/*
exit(0);
assert(expression);  // 断言
cout << flush;  // 刷新缓冲流
string line;
getline(cin, line);
stringstram sline(line);  // string流
cout << unitbuf;  //所有输出操作后都会立即刷新缓冲区
//任何输出都立即刷新，无缓冲
cout << nounitbuf;  //回到正常的缓冲方式
*/
/******************************************************************/

const int INF = 0x3f3f3f3f;
const int N = 2000 + 10;
const int mod = 1e9 + 7;
const long long One = 1LL;

int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};

struct node{
	int x, y, z;
	bool operator < (const node& a) const{
		return x < a.x;
	}
};

string s;
int n, m, k;

int f[N];

void init(){
    for (int i = 0; i < N; i ++)
        f[i] = i;
}

int find(int x){
    if (x != f[x])
        f[x] = find(f[x]);

    return f[x];
}

bool merge(int x, int y){
    int fx = find(x), fy = find(y);
    if (fx == fy)
        return false;

    f[fx] = fy;
    return true;
}

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

    unordered_map<int, vector<pii>> mp;

    for (int i = 1; i <= n; i ++)
        for (int j = i + 1; j <= n; j ++){
            mp[abs(a[i] - a[j])].push_back({i, j});
        }

    vector b(n + 1, vector<pii>());

    for (int i = 1; i < n; i ++){
        for (auto [x, y] : mp){
            if (x % i == 0){
                for (auto j : y){
                    b[i].eb(j);
                    // if (i == 1){
                    //     debug(j.fi, j.se);
                    // }
                }
            }
        }
    }
    for (int i = 1; i < n; i ++){
        if (b[i].size() == 0){
            cout << "No\n";
            return;
        }
    }
    vector<pii> ans(n);
    vector<int> st(n + 1);
    for (int i = n - 1; i; i --){
        for (auto [x, y] : b[i]){
            if (merge(x, y)){
                ans[i] = {x, y};
                break;
            }
        }
    }

    cout << "Yes\n";
    for (int i = 1; i < n; i ++){
        cout << ans[i].fi << " " << ans[i].se << endl;
    }

}

signed main(void){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout << fixed << setprecision(20);

	int t = 1;
	cin >> t;
	while (t --){
        init();
		solve();
	}

	return 0;
}
/*
input:

output:

*/

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to consider scenarios where the code might fail. Here are the key points to consider:

1. **Edge Cases**:
   - Single vertex (n = 1).
   - Two vertices with the same value (n = 2).
   - Two vertices with different values (n = 2).
   - Three vertices with values that do not allow easy pairing (n = 3).

2. **General Cases**:
   - A small graph where the values of `a` are such that it is difficult to find pairs that satisfy the condition for all operations.

3. **Constraints**:
   - The input should be small enough to be manageable but challenging enough to expose the bug.

Let's create a test case that fits these criteria:

- **Test Case 1**: Single vertex.
- **Test Case 2**: Two vertices with the same value.
- **Test Case 3**: Two vertices with different values.
- **Test Case 4**: Three vertices with values that make it difficult to form a connected graph.

Here is the test input:

```plaintext
4
1
10
2
10 10
2
10 11
3
10 12 13
```

Explanation:
1. **Test Case 1**: Single vertex (n = 1).
   - Expected output: "No" (since no operations can be performed).
2. **Test Case 2**: Two vertices with the same value (n = 2).
   - Expected output: "Yes" (since the only operation can connect the two vertices).
3. **Test Case 3**: Two vertices with different values (n = 2).
   - Expected output: "No" (since the only operation cannot connect the two vertices).
4. **Test Case 4**: Three vertices with values that make it difficult to form a connected graph (n = 3).
   - Expected output: "No" (since it is challenging to find pairs that satisfy the condition for all operations).

This test input should help expose the bug in the buggy code by covering a variety of edge cases and small-scale general cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Funny Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Vanya has a graph with $$$n$$$ vertices (numbered from $$$1$$$ to $$$n$$$) and an array $$$a$$$ of $$$n$$$ integers; initially, there are no edges in the graph. Vanya got bored, and to have fun, he decided to perform $$$n - 1$$$ operations.

Operation number $$$x$$$ (operations are numbered in order starting from $$$1$$$) is as follows:

- Choose $$$2$$$ different numbers $$$1 \leq u,v \leq n$$$, such that $$$|a_u - a_v|$$$ is divisible by $$$x$$$.
- Add an undirected edge between vertices $$$u$$$ and $$$v$$$ to the graph.

Help Vanya get a connected$$$^{\text{∗}}$$$ graph using the $$$n - 1$$$ operations, or determine that it is impossible.

Input: Each test consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^{3}$$$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains the number $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of vertices in the graph.

The second line of each test case contains $$$n$$$ numbers $$$a_1, a_2, \cdots a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$.

Output: For each test case, if there is no solution, then output "No" (without quotes).

Otherwise, output "Yes" (without quotes), and then output $$$n - 1$$$ lines, where in the $$$i$$$-th line, output the numbers $$$u$$$ and $$$v$$$ that need to be chosen for operation $$$i$$$.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
2
1 4
4
99 7 1 13
5
10 2 31 44 73
5
87 6 81 44 32
5
62 35 33 79 16
5
6 51 31 69 42
5
52 63 25 21 5
12
33 40 3 11 31 43 37 8 50 5 12 22
output:
YES
2 1
YES
4 1
2 1
3 2
YES
5 1
4 1
3 1
2 1
YES
4 1
3 1
2 1
5 4
YES
3 1
5 1
2 1
4 2
YES
4 1
5 1
2 1
3 2
YES
2 1
5 2
3 1
4 3
YES
9 1
12 9
11 1
10 1
6 1
7 6
2 1
8 2
5 2
3 1
4 1

Note: Let's consider the second test case.

- First operation $$$(x = 1)$$$: we can connect vertices $$$4$$$ and $$$1$$$, since $$$|a_4 - a_1| = |13 - 99| = |-86| = 86$$$, and $$$86$$$ is divisible by $$$1$$$.

- Second operation $$$(x = 2)$$$: we can connect vertices $$$2$$$ and $$$1$$$, since $$$|a_2 - a_1| = |7 - 99| = |-92| = 92$$$, and $$$92$$$ is divisible by $$$2$$$.

- Third operation $$$(x = 3)$$$: we can connect vertices $$$3$$$ and $$$2$$$, since $$$|a_3 - a_2| = |1 - 7| = |-6| = 6$$$, and $$$6$$$ is divisible by $$$3$$$.

Buggy Code:
#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

using namespace std;

struct PairHash {inline std::size_t operator()(const std::pair<long long, long long> &v) const { return v.first * 31ll + v.second; }};

// speed
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);

// alias
using ll = long long;
using ld = long double;
using ull = unsigned long long;

// constants
const ld PI = acosl(-1.0);  /* pi */
const ll INF = 1e18;
const ld EPS = 1e-15;
const ll MAX_N = 202020;
const ll mod = 1e9 + 7;

// typedef
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
typedef array<int,3> ai3;
typedef array<ll,3> all3;
typedef array<ll,4> all4;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all4> vall4;
typedef vector<all5> vall5;
typedef pair<ld, ld> pld;
typedef vector<pld> vpld;
typedef vector<ld> vld;
typedef vector<ll> vll;
typedef vector<ull> vull;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<bool> vb;
typedef deque<ll> dqll;
typedef deque<pll> dqpll;
typedef pair<string, string> pss;
typedef vector<pss> vpss;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<pll, ll, PairHash> umpll;

// macros
#define precision(x) cout<<fixed;cout.precision(x);
#define rep(i,m,n) for(ll i=m;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define uniq(a) sort(all(a)), a.erase(unique(all(a)),end(a))
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define INF(a) memset(a,0x3f3f3f3f3f3f3f3fLL,sizeof(a))
#define NEGINF(a) memset(a,0xcf,sizeof(a))
#define ASCEND(a) iota(all(a),0)
#define sz(x) ll((x).size())
#define BIT(a,i) ((a>>i)&1)
#define BITSHIFT(a,i,n) (((a<<i) & ((1ll<<n) - 1)) | (a>>(n-i)))
#define MAXBIT(a) (64ll - __builtin_clzll(a) - 1ll)
#define MINBIT(a) (__builtin_ctzll(a))
#define pyes cout<<"YES\n";
#define pno cout<<"NO\n";
#define endl "\n"
#define pneg1 cout<<"-1\n";
#define ppossible cout<<"possible\n";
#define pimpossible cout<<"impossible\n";
#define TC(x) cout<<"Case #"<<x<<": ";
#define X first
#define Y second

// debug
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename A>
void __print(const A &x);
template <typename A, typename B>
void __print(const pair<A, B> &p);
template <typename... A>
void __print(const tuple<A...> &t);
template <typename T>
void __print(stack<T> s);
template <typename T>
void __print(queue<T> q);
template <typename T, typename... U>
void __print(priority_queue<T, U...> q);
template <typename A>
void __print(const A &x) {
    bool first = true;
    cerr << '{';
    for (const auto &i : x) {
        cerr << (first ? "" : ","), __print(i);
        first = false;
    }
    cerr << '}';
}
template <typename A, typename B>
void __print(const pair<A, B> &p) {
    cerr << '(';
    __print(p.first);
    cerr << ',';
    __print(p.second);
    cerr << ')';
}
template <typename... A>
void __print(const tuple<A...> &t) {
    bool first = true;
    cerr << '(';
    apply([&first](const auto &...args) { ((cerr << (first ? "" : ","), __print(args), first = false), ...); }, t);
    cerr << ')';
}
template <typename T>
void __print(stack<T> s) {
    vector<T> debugVector;
    while (!s.empty()) {
        T t = s.top();
        debugVector.push_back(t);
        s.pop();
    }
    reverse(debugVector.begin(), debugVector.end());
    __print(debugVector);
}
template <typename T>
void __print(queue<T> q) {
    vector<T> debugVector;
    while (!q.empty()) {
        T t = q.front();
        debugVector.push_back(t);
        q.pop();
    }
    __print(debugVector);
}
template <typename T, typename... U>
void __print(priority_queue<T, U...> q) {
    vector<T> debugVector;
    while (!q.empty()) {
        T t = q.top();
        debugVector.push_back(t);
        q.pop();
    }
    __print(debugVector);
}
void _print() { cerr << "]\n"; }
template <typename Head, typename... Tail>
void _print(const Head &H, const Tail &...T) {
    __print(H);
    if (sizeof...(T))
        cerr << ", ";
    _print(T...);
}
#ifndef ONLINE_JUDGE
#define debug(...) cerr << "Line:" << __LINE__ << " [" << #__VA_ARGS__ << "] = ["; _print(__VA_ARGS__);
#else
#define debug(...)
#endif

// utility functions
template <typename T>
void print(T &&t)  { cout << t << "\n"; }
template<typename T>
void printv(vector<T>v){ll n=v.size();rep(i,0,n){cout<<v[i];if(i+1!=n)cout<<' ';}cout<<endl;}
template<typename T>
void printvv(vector<vector<T>>v){ll n=v.size();rep(i,0,n)printv(v[i]);}
template<typename T>
void printvln(vector<T>v){ll n=v.size();rep(i,0,n)cout<<v[i]<<endl;}
void fileIO(string in = "input.txt", string out = "output.txt") {freopen(in.c_str(),"r",stdin); freopen(out.c_str(),"w",stdout);}
void hackercupIO(string in) {fileIO("/Users/hayoungsong/Downloads/" + in + ".txt", "/Users/hayoungsong/Downloads/" + in + "_solution.txt");}
void readf() {freopen("", "rt", stdin);}
template <typename... T>
void in(T &...a) { ((cin >> a), ...); }
template<typename T>
void in(vector<T>& v){rep(i,0,sz(v)) in(v[i]);}
template<typename T>
void in(deque<T>& v){rep(i,0,sz(v)) in(v[i]);}
template<typename T, typename U>
void in(pair<T,U>& A) {in(A.first, A.second);}
template<typename T, typename U>
void in(vector<pair<T,U>>& A) {rep(i,0,sz(A)) in(A[i]); }
template<typename T, std::size_t N>
void in(vector<array<T,N>>& A) {rep(i,0,sz(A)) rep(j,0,sz(A[i])) in(A[i][j]); }
template<typename T>
void in(vector<vector<T>>& A) {rep(i,0,sz(A)) in(A[i]);}

struct Combination {
    vll fac, inv;
    ll n, MOD;

    ll modpow(ll n, ll x, ll MOD = mod) { if(!x) return 1; ll res = modpow(n,x>>1,MOD); res = (res * res) % MOD; if(x&1) res = (res * n) % MOD; return res; }

    Combination(ll _n, ll MOD = mod): n(_n + 1), MOD(MOD) {
        inv = fac = vll(n,1);
        rep(i,1,n) fac[i] = fac[i-1] * i % MOD;
        inv[n - 1] = modpow(fac[n - 1], MOD - 2, MOD);
        rrep(i,1,n - 2) inv[i] = inv[i + 1] * (i + 1) % MOD;
    }

    ll fact(ll n) {return fac[n];}
    ll nCr(ll n, ll r) {
        if(n < r or n < 0 or r < 0) return 0;
        return fac[n] * inv[r] % MOD * inv[n-r] % MOD;
    }
};

struct Matrix {
    ll r,c,MOD;
    vvll matrix;
    Matrix(ll r, ll c, ll v = 0, ll MOD = mod): r(r), c(c), matrix(vvll(r,vll(c,v))), MOD(MOD) {}
    Matrix(vvll m, ll MOD = mod) : r(sz(m)), c(sz(m[0])), matrix(m), MOD(MOD) {}

    vector<ll>& operator[](ll pos) {return matrix[pos];}
    Matrix operator*(const Matrix& B) const {
        Matrix res(r, B.c, 0,MOD);
        rep(i,0,r) rep(j,0,B.c) rep(k,0,B.r) {
                    res[i][j] = (res[i][j] + matrix[i][k] * B.matrix[k][j] % MOD) % MOD;
                }
        return res;
    }

    Matrix copy() {
        Matrix copy(r,c,0,MOD);
        copy.matrix = matrix;
        return copy;
    }

    Matrix pow(ll n) {
        assert(r == c);
        Matrix res(r,r, 0,MOD);
        Matrix now = copy();
        rep(i,0,r) res[i][i] = 1;
        while(n) {
            if(n & 1) res = res * now;
            now = now * now;
            n /= 2;
        }
        return res;
    }

    ll det() {
        if(r == 1) return matrix[0][0];
        if(r == 2) return matrix[0][0] * matrix[1][1] - matrix[0][1] * matrix[1][0];
        ll res = 0;
        rep(p,0,c) {
            Matrix mat(c-1,c-1);
            rep(i,1,r) rep(j,0,c) {
                    if(j == p) continue;
                    mat[i][j - (j >= p)] = matrix[i][j];
                }
            res += matrix[0][p] * (p & 1 ? -1 : 1) * mat.det();
        }
        return res;
    }
};

// geometry data structures
template <typename T>
struct Point {
    T y,x;
    Point(T y, T x) : y(y), x(x) {}
    Point(pair<T,T> p) : y(p.first), x(p.second) {}
    Point() {}
    void input() {cin>>y>>x;}
    friend ostream& operator<<(ostream& os, const Point<T>& p) { os<<p.y<<' '<<p.x<<'\n'; return os;}
    Point<T> operator+(Point<T>& p) {return Point<T>(y + p.y, x + p.x);}
    Point<T> operator-(Point<T>& p) {return Point<T>(y - p.y, x - p.x);}
    Point<T> operator*(ll n) {return Point<T>(y*n,x*n); }
    Point<T> operator/(ll n) {return Point<T>(y/n,x/n); }
    bool operator==(const Point<T> &a) {return x == a.x && y == a.y;}
    bool operator!=(const Point<T> &a) {return x != a.x or y != a.y;}
    bool operator<(const Point &other) const {if (x == other.x) return y < other.y;return x < other.x;}
    Point<T> rotate(Point<T> center, ld angle) {
        ld si = sin(angle * PI / 180.), co = cos(angle * PI / 180.);
        ld y = this->y - center.y;
        ld x = this->x - center.x;

        return Point<T>(y * co - x * si + center.y, y * si + x * co + center.x);
    }
    ld distance(Point<T> other) {
        T dy = abs(this->y - other.y);
        T dx = abs(this->x - other.x);
        return sqrt(dy * dy + dx * dx);
    }

    T norm() { return x * x + y * y; }
};

template<typename T>
struct Line {
    Point<T> A, B;
    Line(Point<T> A, Point<T> B) : A(A), B(B) {}
    Line() {}

    void input() {
        A = Point<T>();
        B = Point<T>();
        A.input();
        B.input();
    }

    T ccw(Point<T> &a, Point<T> &b, Point<T> &c) {
        T res = a.x * b.y + b.x * c.y + c.x * a.y;
        res -= (a.x * c.y + b.x * a.y + c.x * b.y);
        return res;
    }

    bool on(Point<T> x) {
        return ccw(A,x,B) == 0;
    }

    bool isIntersect(Line<T> o) {
        T p1p2 = ccw(A,B,o.A) * ccw(A,B,o.B);
        T p3p4 = ccw(o.A,o.B,A) * ccw(o.A,o.B,B);
        if (p1p2 == 0 && p3p4 == 0) {
            pair<T,T> p1(A.y, A.x), p2(B.y,B.x), p3(o.A.y, o.A.x), p4(o.B.y, o.B.x);
            if (p1 > p2) swap(p2, p1);
            if (p3 > p4) swap(p3, p4);
            return p3 <= p2 && p1 <= p4;
        }
        return p1p2 <= 0 && p3p4 <= 0;
    }

    pair<bool,Point<ld>> intersection(Line<T> o) {
        if(!this->intersection(o)) return {false, {}};
        ld det = 1. * (o.B.y-o.A.y)*(B.x-A.x) - 1.*(o.B.x-o.A.x)*(B.y-A.y);
        ld t = ((o.B.x-o.A.x)*(A.y-o.A.y) - (o.B.y-o.A.y)*(A.x-o.A.x)) / det;
        return {true, {A.y + 1. * t * (B.y - A.y), B.x + 1. * t * (B.x - A.x)}};
    }

    //@formula for : y = ax + pre
    //@return {a,pre};
    pair<ld, ld> formula() {
        T y1 = A.y, y2 = B.y;
        T x1 = A.x, x2 = B.x;
        if(y1 == y2) return {0, (ld)y1};
        if(x1 == x2) return {INF, (ld)x1};
        ld a = 1. * (y2 - y1) / (x2 - x1);
        ld b = -x1 * a + y1;
        return {a, b};
    }
};

template<typename T>
struct Circle {
    Point<T> center;
    T radius;
    Circle(T y, T x, T radius) : center(Point<T>(y,x)), radius(radius) {}
    Circle(Point<T> center, T radius) : center(center), radius(radius) {}
    Circle() {}

    void input() {
        center = Point<T>();
        center.input();
        cin>>radius;
    }

    bool circumference(Point<T> p) {
        return (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y) == radius * radius;
    }

    bool intersect(Circle<T> c) {
        T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
        return (radius - c.radius) * (radius - c.radius) <= d and d <= (radius + c.radius) * (radius + c.radius);
    }

    bool include(Circle<T> c) {
        T d = (center.x - c.center.x) * (center.x - c.center.x) + (center.y - c.center.y) * (center.y - c.center.y);
        return d <= radius * radius;
    }

    bool include(Point<T> p) {
        T d = (center.x - p.x) * (center.x - p.x) + (center.y - p.y) * (center.y - p.y);
        return d <= radius * radius;
    }
};

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
all3 __exgcd(ll x, ll y) { if(!y) return {x,1,0}; auto [g,x1,y1] = __exgcd(y, x % y); return {g, y1, x1 - (x/y) * y1}; }
ll __lcm(ll x, ll y) { return x / __gcd(x,y) * y; }
ll modpow(ll n, ll x, ll MOD = mod) {if(x<0){return modpow(modpow(n,-x,MOD),MOD-2,MOD);}n%=MOD;ll res=1;while(x){if(x&1){res=res*n%MOD;}n=n*n%MOD;x>>=1;}return res;}
ll __xor(ll n) {return n%4==0?n:n%4==1?1:n%4==2?n+1:0;}
ll __rangexor(ll l, ll r) {return __xor(r)^__xor(l-1);}

ll uf[2020];
ll find(ll u) {
    return uf[u] == u ? u : uf[u] = find(uf[u]);
}
void uni(ll u, ll v) {
    ll pu = find(u), pv = find(v);
    uf[pu] = uf[pv] = min(pu,pv);
}
pair<bool,vvll> solve(vll& A, ll n) {
    ASCEND(uf);
    vector<vall3> adj(n);
    vall3 C;
    rep(i,0,n) rep(j,i+1,n) {
        ll d = abs(A[i] - A[j]);
        if(!d) C.push_back({0,i+1,j+1});
        else if(d <= n -1) adj[d].push_back({d,i+1,j+1});
        else {
            rrep(k,1,n-1) {
                if(d % k) continue;
                adj[k].push_back({d,i+1,j+1});
                break;
            }
        }
    }
    vvll res;
    rrep(i,1,n-1) {
        bool ok = false;
        vall3 B;
        while(sz(adj[i])) {
            auto [d,u,v] = adj[i].back(); adj[i].pop_back();
            if(find(u) == find(v)) continue;
            if(ok) {
                B.push_back({d,u,v});
            } else {
                ok = true;
                uni(u,v);
                res.push_back({u,v});
            }
        }
        for(auto& [d,u,v] : C) {
            if(find(u) == find(v)) continue;
            if(d % i or ok) B.push_back({d,u,v});
            else {
                ok = true;
                uni(u,v);
                res.push_back({u,v});
            }
        }
        if(!ok) return {false, {}};
        swap(C,B);
    }
    reverse(all(res));
    return {true,res};
}
int main() {
    Code By Sumfi
    precision(18)
    ll tc = 1;
    in(tc);
    rep(i,1,tc+1) {
        ll n;
        in(n);
        vll A(n);
        in(A);
        auto [ok,res] = solve(A,n);
        if(ok) {
            pyes
            printvv(res);
        } else pno
    }
    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 generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that might cause the algorithm to fail. Here are the steps to think through this:

1. **Identify Potential Issues:**
   - The code needs to ensure that for each operation, there are valid pairs of vertices whose difference is divisible by the operation number.
   - The code must also ensure that the graph remains connected after each operation.

2. **Edge Cases:**
   - **Single Element Array:** This is a trivial case where the graph cannot be connected.
   - **Small Arrays with Non-Divisible Differences:** This can test the ability of the algorithm to find valid pairs.
   - **Large Differences:** This can test the ability to handle large numbers and ensure the differences are divisible by the operation number.
   - **Equal Elements:** This can test the ability to handle cases where all elements are the same, which should trivially allow connections.

3. **Construct the Test Case:**
   - For a small-scale test input, we can use an array of 5 elements to ensure the test case is valid and manageable.
   - We will use a mix of small and large numbers to ensure the algorithm is robust.

Given these considerations, here is a small-scale test input that should be challenging for the buggy code:

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

### Explanation:
- **Number of Test Cases:** 1
- **Number of Vertices:** 5
- **Array of Elements:** [1, 2, 3, 4, 5]

This test case is designed to:
- Ensure that the differences between the elements are small and manageable.
- Test the ability to find valid pairs for each operation (1 to 4).
- Ensure that the graph remains connected after each operation.

This input should be sufficient to expose any issues in the buggy code, especially in handling small differences and ensuring the graph remains connected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Funny Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Vanya has a graph with $$$n$$$ vertices (numbered from $$$1$$$ to $$$n$$$) and an array $$$a$$$ of $$$n$$$ integers; initially, there are no edges in the graph. Vanya got bored, and to have fun, he decided to perform $$$n - 1$$$ operations.

Operation number $$$x$$$ (operations are numbered in order starting from $$$1$$$) is as follows:

- Choose $$$2$$$ different numbers $$$1 \leq u,v \leq n$$$, such that $$$|a_u - a_v|$$$ is divisible by $$$x$$$.
- Add an undirected edge between vertices $$$u$$$ and $$$v$$$ to the graph.

Help Vanya get a connected$$$^{\text{∗}}$$$ graph using the $$$n - 1$$$ operations, or determine that it is impossible.

Input: Each test consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^{3}$$$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains the number $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of vertices in the graph.

The second line of each test case contains $$$n$$$ numbers $$$a_1, a_2, \cdots a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$.

Output: For each test case, if there is no solution, then output "No" (without quotes).

Otherwise, output "Yes" (without quotes), and then output $$$n - 1$$$ lines, where in the $$$i$$$-th line, output the numbers $$$u$$$ and $$$v$$$ that need to be chosen for operation $$$i$$$.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
2
1 4
4
99 7 1 13
5
10 2 31 44 73
5
87 6 81 44 32
5
62 35 33 79 16
5
6 51 31 69 42
5
52 63 25 21 5
12
33 40 3 11 31 43 37 8 50 5 12 22
output:
YES
2 1
YES
4 1
2 1
3 2
YES
5 1
4 1
3 1
2 1
YES
4 1
3 1
2 1
5 4
YES
3 1
5 1
2 1
4 2
YES
4 1
5 1
2 1
3 2
YES
2 1
5 2
3 1
4 3
YES
9 1
12 9
11 1
10 1
6 1
7 6
2 1
8 2
5 2
3 1
4 1

Note: Let's consider the second test case.

- First operation $$$(x = 1)$$$: we can connect vertices $$$4$$$ and $$$1$$$, since $$$|a_4 - a_1| = |13 - 99| = |-86| = 86$$$, and $$$86$$$ is divisible by $$$1$$$.

- Second operation $$$(x = 2)$$$: we can connect vertices $$$2$$$ and $$$1$$$, since $$$|a_2 - a_1| = |7 - 99| = |-92| = 92$$$, and $$$92$$$ is divisible by $$$2$$$.

- Third operation $$$(x = 3)$$$: we can connect vertices $$$3$$$ and $$$2$$$, since $$$|a_3 - a_2| = |1 - 7| = |-6| = 6$$$, and $$$6$$$ is divisible by $$$3$$$.

Buggy Code:
# for I/O for local system
import sys
from os import path
if(path.exists('Input.txt')):
    sys.stdin = open('Input.txt','r')
    sys.stdout = open('Output.txt','w')

# For fast I/O
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
print = sys.stdout.write

# Import libraries here whenever required
from collections import *
from random import randint

# Use this because normal dict can sometimes give TLE
class mydict:
    def __init__(self, func=lambda: 0):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}
    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]
    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]
    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item
    def getkeys(self):
        return [self.random ^ i for i in self.dict]
    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'

class DSU:
    def __init__(self, n):
        self.parent = [*range(n+1)]
        self.size = [1]*(n+1)
        self.min, self.max = [*range(n+1)], [*range(n+1)]
        self.count = n
    def find(self, a):
        if self.parent[a] == a:
            return a
        x = a
        while a != self.parent[a]:
            a = self.parent[a]
        while x != self.parent[x]:
            self.parent[x], x = a, self.parent[x]
        return a
    def union(self, a, b):
        a, b = self.find(a), self.find(b)
        if a != b:
            if self.size[a] > self.size[b]:
                a, b = b, a
            # self.parent[a] = max(self.parent[a],self.parent[b])
            # self.parent[b] = max(self.parent[a],self.parent[b])
            self.parent[a] = b
            self.size[b] += self.size[a]
            self.min[b] = min(self.min[a], self.min[b])
            self.max[b] = max(self.max[a], self.max[b])
            self.count -= 1
    def countSets(self):
        return self.count

def yo():
    dsu = DSU(n + 5)
    paired = [[0 for i in range(n)] for i in range(n)]
    cnt = [0] * n
    for_x = [[] for i in range(n)]
    xer = [0] * n
    adj = [[] for i in range(n)]
    vis = [False] * n
    for_ans = [[] for i in range(n)]
    for i in range(2, n):
        diff = [[] for ii in range(i)]
        for j in range(n):
            rem = a[j] % i
            diff[rem].append(j)
        for rem in range(i):
            if(len(diff[rem]) <= 1):
                continue
            for val in diff[rem]:
                cnt[val] += 1
                for_x[val].append(i)
            xer[i] += len(diff[rem])
        if(xer[i] == 2):
            cc = []
            for rem in range(i):
                if(len(diff[rem]) <= 1):
                    continue
                for val in diff[rem]:
                    cnt[val] += 1
                    for_x[val].append(i)
                    cc.append(val)
            if(dsu.find(cc[0]) != dsu.find(cc[1])):
                dsu.union(cc[0], cc[1])
                adj[cc[0]].append(cc[1])
                adj[cc[1]].append(cc[0])
                vis[i] = True
                for_ans[i] = [cc[0], cc[1]]
    mn = [10 ** 8] * n
    l = []
    for i in range(n):
        l.append([cnt[i], i])
    l.sort()
    # print(l)
    vis = [False] * n
    for iiiii in range(2):
        for i in range(n):
            c, idx = l[i]
            if(c == 0):
                continue
            for_x[idx]
            nfor_x = []
            for xx in for_x[idx]:
                nfor_x.append([xer[xx], xx])
            nfor_x.sort()
            for tmp, xx in nfor_x:
                if(vis[xx]):
                    continue
                u = idx
                v = -1
                for j in range(n):
                    if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
                        if(a[j] % xx == a[idx] % xx):
                            v = j
                            break
                if(v != -1):
                    vis[xx] = True
                    paired[u][v] = paired[v][u] = 1
                    for_ans[xx] = [u, v]
                    adj[u].append(v)
                    adj[v].append(u)
                    dsu.union(u, v)
                    break
                else:
                    continue
    comps = 0
    # print(vis)
    # print(for_ans)
    vis = [0] * n
    helper = []
    # print(adj)
    for i in range(n):
        if(vis[i]):
            continue
        helper.append(i)
        q = deque()
        q.append(i)
        vis[i] = 1
        comps += 1
        if(comps > 2):
            # print(-1)
            # print(n)
            # print(*a)
            # print(-1)
            # assert 1 == 0
            # print("No\n")
            return []
        while q:
            k = len(q)
            for j in range(k):
                node = q.popleft()
                for nbr in adj[node]:
                    if(not vis[nbr]):
                        vis[nbr] = 1
                        q.append(nbr)
    if(comps != 2):
        # assert 1 == 0
        # print("No\n")
        return []
    for_ans[1] = [helper[0], helper[1]]
    return for_ans

def yoyo():
    dsu = DSU(n + 5)
    paired = [[0 for i in range(n)] for i in range(n)]
    cnt = [0] * n
    for_x = [[] for i in range(n)]
    xer = [0] * n
    adj = [[] for i in range(n)]
    vis = [False] * n
    for_ans = [[] for i in range(n)]
    for i in range(2, n):
        diff = [[] for ii in range(i)]
        for j in range(n):
            rem = a[j] % i
            diff[rem].append(j)
        for rem in range(i):
            if(len(diff[rem]) <= 1):
                continue
            for val in diff[rem]:
                cnt[val] += 1
                for_x[val].append(i)
            xer[i] += len(diff[rem])
        if(xer[i] == 2):
            cc = []
            for rem in range(i):
                if(len(diff[rem]) <= 1):
                    continue
                for val in diff[rem]:
                    cnt[val] += 1
                    for_x[val].append(i)
                    cc.append(val)
            if(dsu.find(cc[0]) != dsu.find(cc[1])):
                dsu.union(cc[0], cc[1])
                adj[cc[0]].append(cc[1])
                adj[cc[1]].append(cc[0])
                vis[i] = True
                for_ans[i] = [cc[0], cc[1]]
    mn = [10 ** 8] * n
    l = []
    for i in range(n):
        l.append([cnt[i], i])
    l.sort()
    # print(l)
    for iiiii in range(2):
        for i in range(n):
            c, idx = l[i]
            if(c == 0):
                continue
            for_x[idx]
            nfor_x = []
            for xx in for_x[idx]:
                nfor_x.append([xer[xx], xx])
            nfor_x.sort()
            once = 0
            for tmp, xx in nfor_x:
                if(vis[xx]):
                    continue
                if(iiiii == 0 and once == 0):
                    once = 1
                    continue
                once = 1
                u = idx
                v = -1
                for j in range(n):
                    if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
                        if(a[j] % xx == a[idx] % xx):
                            v = j
                            break
                if(v != -1):
                    vis[xx] = True
                    paired[u][v] = paired[v][u] = 1
                    for_ans[xx] = [u, v]
                    adj[u].append(v)
                    adj[v].append(u)
                    dsu.union(u, v)
                    break
                else:
                    continue
    comps = 0
    # print(vis)
    vis = [0] * n
    helper = []
    # print(adj)
    for i in range(n):
        if(vis[i]):
            continue
        helper.append(i)
        q = deque()
        q.append(i)
        vis[i] = 1
        comps += 1
        if(comps > 2):
            # print(-1)
            # print(n)
            # print(*a)
            # print(-1)
            # assert 1 == 0
            # print("No\n")
            return []
        while q:
            k = len(q)
            for j in range(k):
                node = q.popleft()
                for nbr in adj[node]:
                    if(not vis[nbr]):
                        vis[nbr] = 1
                        q.append(nbr)
    if(comps != 2):
        # assert 1 == 0
        # print("No\n")
        return []
    for_ans[1] = [helper[0], helper[1]]
    return for_ans

def yoyoyo():
    dsu = DSU(n + 5)
    paired = [[0 for i in range(n)] for i in range(n)]
    cnt = [0] * n
    for_x = [[] for i in range(n)]
    xer = [0] * n
    adj = [[] for i in range(n)]
    vis = [False] * n
    for_ans = [[] for i in range(n)]
    for i in range(2, n):
        diff = [[] for ii in range(i)]
        for j in range(n):
            rem = a[j] % i
            diff[rem].append(j)
        for rem in range(i):
            if(len(diff[rem]) <= 1):
                continue
            for val in diff[rem]:
                cnt[val] += 1
                for_x[val].append(i)
            xer[i] += len(diff[rem])
        if(xer[i] == 2):
            cc = []
            for rem in range(i):
                if(len(diff[rem]) <= 1):
                    continue
                for val in diff[rem]:
                    cnt[val] += 1
                    for_x[val].append(i)
                    cc.append(val)
            if(dsu.find(cc[0]) != dsu.find(cc[1])):
                dsu.union(cc[0], cc[1])
                adj[cc[0]].append(cc[1])
                adj[cc[1]].append(cc[0])
                vis[i] = True
                for_ans[i] = [cc[0], cc[1]]
    mn = [10 ** 8] * n
    l = []
    for i in range(n):
        l.append([cnt[i], i])
    l.sort()
    # print(l)
    for iiiii in range(2):
        for i in range(n):
            c, idx = l[i]
            if(c == 0):
                continue
            for_x[idx]
            nfor_x = []
            for xx in for_x[idx]:
                nfor_x.append([xer[xx], xx])
            nfor_x.sort()
            for tmp, xx in nfor_x:
                if(vis[xx]):
                    continue
                if(iiiii < 1):
                    if(randint(0, 1000) % 2 == 0):
                        continue
                u = idx
                v = -1
                for j in range(n):
                    if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
                        if(a[j] % xx == a[idx] % xx):
                            v = j
                            break
                if(v != -1):
                    vis[xx] = True
                    paired[u][v] = paired[v][u] = 1
                    for_ans[xx] = [u, v]
                    adj[u].append(v)
                    adj[v].append(u)
                    dsu.union(u, v)
                    break
                else:
                    continue
    comps = 0
    # print(vis)
    vis = [0] * n
    helper = []
    # print(adj)
    for i in range(n):
        if(vis[i]):
            continue
        helper.append(i)
        q = deque()
        q.append(i)
        vis[i] = 1
        comps += 1
        if(comps > 2):
            # print(-1)
            # print(n)
            # print(*a)
            # print(-1)
            # assert 1 == 0
            # print("No\n")
            return []
        while q:
            k = len(q)
            for j in range(k):
                node = q.popleft()
                for nbr in adj[node]:
                    if(not vis[nbr]):
                        vis[nbr] = 1
                        q.append(nbr)
    if(comps != 2):
        # assert 1 == 0
        # print("No\n")
        return []
    for_ans[1] = [helper[0], helper[1]]
    return for_ans

# Solver function
def solve():
    if(n == 1):
        print("Yes\n")
        return
    # dsu = DSU(n + 5)
    # paired = [[0 for i in range(n)] for i in range(n)]
    # cnt = [0] * n
    # for_x = [[] for i in range(n)]
    # xer = [0] * n
    # adj = [[] for i in range(n)]
    # vis = [False] * n
    # for i in range(2, n):
    #     diff = [[] for ii in range(i)]
    #     for j in range(n):
    #         rem = a[j] % i
    #         diff[rem].append(j)
    #     for rem in range(i):
    #         if(len(diff[rem]) <= 1):
    #             continue
    #         for val in diff[rem]:
    #             cnt[val] += 1
    #             for_x[val].append(i)
    #         xer[i] += len(diff[rem])
    #     if(xer[i] == 2):
    #         cc = []
    #         for rem in range(i):
    #             if(len(diff[rem]) <= 1):
    #                 continue
    #             for val in diff[rem]:
    #                 cnt[val] += 1
    #                 for_x[val].append(i)
    #                 cc.append(val)
    #         if(dsu.find(cc[0]) != dsu.find(cc[1])):
    #             dsu.union(cc[0], cc[1])
    #             adj[cc[0]].append(cc[1])
    #             adj[cc[1]].append(cc[0])
    #             vis[i] = True
    # mn = [10 ** 8] * n
    # l = []
    # for i in range(n):
    #     l.append([cnt[i], i])
    # l.sort()
    # for_ans = [[] for i in range(n)]
    # # print(l)
    # for iiiii in range(2):
    #     for i in range(n):
    #         c, idx = l[i]
    #         if(c == 0):
    #             continue
    #         for_x[idx]
    #         nfor_x = []
    #         for xx in for_x[idx]:
    #             nfor_x.append([xer[xx], xx])
    #         nfor_x.sort()
    #         for tmp, xx in nfor_x:
    #             if(vis[xx]):
    #                 continue
    #             u = idx
    #             v = -1
    #             for j in range(n):
    #                 if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
    #                     if(a[j] % xx == a[idx] % xx):
    #                         v = j
    #                         break
    #             if(v != -1):
    #                 vis[xx] = True
    #                 paired[u][v] = paired[v][u] = 1
    #                 for_ans[xx] = [u, v]
    #                 adj[u].append(v)
    #                 adj[v].append(u)
    #                 dsu.union(u, v)
    #                 break
    #             else:
    #                 continue
    # comps = 0
    # # print(vis)
    # vis = [0] * n
    # helper = []
    # # print(adj)
    # for i in range(n):
    #     if(vis[i]):
    #         continue
    #     helper.append(i)
    #     q = deque()
    #     q.append(i)
    #     vis[i] = 1
    #     comps += 1
    #     if(comps > 2):
    #         # print(-1)
    #         # print(n)
    #         # print(*a)
    #         # print(-1)
    #         # assert 1 == 0
    #         print("No\n")
    #         return
    #     while q:
    #         k = len(q)
    #         for j in range(k):
    #             node = q.popleft()
    #             for nbr in adj[node]:
    #                 if(not vis[nbr]):
    #                     vis[nbr] = 1
    #                     q.append(nbr)
    # if(comps != 2):
    #     # assert 1 == 0
    #     print("No\n")
    #     return
    # for_ans[1] = [helper[0], helper[1]]
    for_ans = yo()
    if(len(for_ans) == 0):
        for_ans = yoyo()
    if(len(for_ans) == 0):
        # assert 1 == 0
        print("No\n")
        return
    if(len(for_ans) == 0):
        for_ans = yoyoyo()
    if(len(for_ans) == 0):
        # assert 1 == 0
        print("No\n")
        return
    print("Yes\n")
    for i in range(1, n):
        u, v = for_ans[i]
        print(str(u + 1) + " " + str(v + 1) + "\n")
        assert (abs(a[u] - a[v]) % i) == 0

# Main
for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    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 the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to think through this:

1. **Edge Cases for Small `n`**:
   - Test with `n = 1` to ensure the code handles the trivial case correctly.
   - Test with `n = 2` to check if the code can handle the smallest non-trivial case.

2. **Edge Cases for Large `n`**:
   - Test with `n` close to the upper limit (2000) to check if the code handles large inputs efficiently.

3. **Special Values for `a`**:
   - Test with all elements in `a` being the same to check if the code can handle uniform values.
   - Test with `a` values that are carefully chosen to maximize the number of operations that need to be performed.

4. **Random Values for `a`**:
   - Test with random values to check the general robustness of the code.

5. **Specific Values for `a`**:
   - Test with values that ensure certain operations fail to find valid pairs, which can help expose bugs in the logic.

Given the constraints and the need for a small-scale test input, let's focus on a test case with `n = 5` and carefully chosen `a` values.

### Chosen Test Case
- **n = 5**
- **a = [1, 2, 3, 4, 5]**

This test case is small enough to be manageable and ensures that the values of `a` are sequential, which can make it challenging for the code to find valid pairs for each operation.

### Output
```plaintext
1
5
1 2 3 4 5
```

This test input will help expose any issues in the logic for finding valid pairs and ensuring the graph remains connected.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Funny Game
time_limit_ms: 2000
memory_limit_mb: 256
Description: Vanya has a graph with $$$n$$$ vertices (numbered from $$$1$$$ to $$$n$$$) and an array $$$a$$$ of $$$n$$$ integers; initially, there are no edges in the graph. Vanya got bored, and to have fun, he decided to perform $$$n - 1$$$ operations.

Operation number $$$x$$$ (operations are numbered in order starting from $$$1$$$) is as follows:

- Choose $$$2$$$ different numbers $$$1 \leq u,v \leq n$$$, such that $$$|a_u - a_v|$$$ is divisible by $$$x$$$.
- Add an undirected edge between vertices $$$u$$$ and $$$v$$$ to the graph.

Help Vanya get a connected$$$^{\text{∗}}$$$ graph using the $$$n - 1$$$ operations, or determine that it is impossible.

Input: Each test consists of multiple test cases. The first line contains an integer $$$t$$$ ($$$1 \le t \le 10^{3}$$$) — the number of test cases. Then follows the description of the test cases.

The first line of each test case contains the number $$$n$$$ ($$$1 \leq n \leq 2000$$$) — the number of vertices in the graph.

The second line of each test case contains $$$n$$$ numbers $$$a_1, a_2, \cdots a_n$$$ ($$$1 \leq a_i \leq 10^9$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2000$$$.

Output: For each test case, if there is no solution, then output "No" (without quotes).

Otherwise, output "Yes" (without quotes), and then output $$$n - 1$$$ lines, where in the $$$i$$$-th line, output the numbers $$$u$$$ and $$$v$$$ that need to be chosen for operation $$$i$$$.

You can output each letter in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Examples:
input:
8
2
1 4
4
99 7 1 13
5
10 2 31 44 73
5
87 6 81 44 32
5
62 35 33 79 16
5
6 51 31 69 42
5
52 63 25 21 5
12
33 40 3 11 31 43 37 8 50 5 12 22
output:
YES
2 1
YES
4 1
2 1
3 2
YES
5 1
4 1
3 1
2 1
YES
4 1
3 1
2 1
5 4
YES
3 1
5 1
2 1
4 2
YES
4 1
5 1
2 1
3 2
YES
2 1
5 2
3 1
4 3
YES
9 1
12 9
11 1
10 1
6 1
7 6
2 1
8 2
5 2
3 1
4 1

Note: Let's consider the second test case.

- First operation $$$(x = 1)$$$: we can connect vertices $$$4$$$ and $$$1$$$, since $$$|a_4 - a_1| = |13 - 99| = |-86| = 86$$$, and $$$86$$$ is divisible by $$$1$$$.

- Second operation $$$(x = 2)$$$: we can connect vertices $$$2$$$ and $$$1$$$, since $$$|a_2 - a_1| = |7 - 99| = |-92| = 92$$$, and $$$92$$$ is divisible by $$$2$$$.

- Third operation $$$(x = 3)$$$: we can connect vertices $$$3$$$ and $$$2$$$, since $$$|a_3 - a_2| = |1 - 7| = |-6| = 6$$$, and $$$6$$$ is divisible by $$$3$$$.

Buggy Code:
# for I/O for local system
import sys
from os import path
if(path.exists('Input.txt')):
    sys.stdin = open('Input.txt','r')
    sys.stdout = open('Output.txt','w')

# For fast I/O
input = sys.stdin.buffer.readline
# input = sys.stdin.readline
print = sys.stdout.write

# Import libraries here whenever required
from collections import *
from random import randint

# Use this because normal dict can sometimes give TLE
class mydict:
    def __init__(self, func=lambda: 0):
        self.random = randint(0, 1 << 32)
        self.default = func
        self.dict = {}
    def __getitem__(self, key):
        mykey = self.random ^ key
        if mykey not in self.dict:
            self.dict[mykey] = self.default()
        return self.dict[mykey]
    def get(self, key, default):
        mykey = self.random ^ key
        if mykey not in self.dict:
            return default
        return self.dict[mykey]
    def __setitem__(self, key, item):
        mykey = self.random ^ key
        self.dict[mykey] = item
    def getkeys(self):
        return [self.random ^ i for i in self.dict]
    def __str__(self):
        return f'{[(self.random ^ i, self.dict[i]) for i in self.dict]}'

class DSU:
    def __init__(self, n):
        self.parent = [*range(n+1)]
        self.size = [1]*(n+1)
        self.min, self.max = [*range(n+1)], [*range(n+1)]
        self.count = n
    def find(self, a):
        if self.parent[a] == a:
            return a
        x = a
        while a != self.parent[a]:
            a = self.parent[a]
        while x != self.parent[x]:
            self.parent[x], x = a, self.parent[x]
        return a
    def union(self, a, b):
        a, b = self.find(a), self.find(b)
        if a != b:
            if self.size[a] > self.size[b]:
                a, b = b, a
            # self.parent[a] = max(self.parent[a],self.parent[b])
            # self.parent[b] = max(self.parent[a],self.parent[b])
            self.parent[a] = b
            self.size[b] += self.size[a]
            self.min[b] = min(self.min[a], self.min[b])
            self.max[b] = max(self.max[a], self.max[b])
            self.count -= 1
    def countSets(self):
        return self.count

def yo():
    dsu = DSU(n + 5)
    paired = [[0 for i in range(n)] for i in range(n)]
    cnt = [0] * n
    for_x = [[] for i in range(n)]
    xer = [0] * n
    adj = [[] for i in range(n)]
    vis = [False] * n
    for_ans = [[] for i in range(n)]
    for i in range(2, n):
        diff = [[] for ii in range(i)]
        for j in range(n):
            rem = a[j] % i
            diff[rem].append(j)
        for rem in range(i):
            if(len(diff[rem]) <= 1):
                continue
            for val in diff[rem]:
                cnt[val] += 1
                for_x[val].append(i)
            xer[i] += len(diff[rem])
        if(xer[i] == 2):
            cc = []
            for rem in range(i):
                if(len(diff[rem]) <= 1):
                    continue
                for val in diff[rem]:
                    cnt[val] += 1
                    for_x[val].append(i)
                    cc.append(val)
            if(dsu.find(cc[0]) != dsu.find(cc[1])):
                dsu.union(cc[0], cc[1])
                adj[cc[0]].append(cc[1])
                adj[cc[1]].append(cc[0])
                vis[i] = True
                for_ans[i] = [cc[0], cc[1]]
    mn = [10 ** 8] * n
    l = []
    for i in range(n):
        l.append([cnt[i], i])
    l.sort()
    # print(l)
    vis = [False] * n
    for iiiii in range(2):
        for i in range(n):
            c, idx = l[i]
            if(c == 0):
                continue
            for_x[idx]
            nfor_x = []
            for xx in for_x[idx]:
                nfor_x.append([xer[xx], xx])
            nfor_x.sort()
            for tmp, xx in nfor_x:
                if(vis[xx]):
                    continue
                u = idx
                v = -1
                for j in range(n):
                    if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
                        if(a[j] % xx == a[idx] % xx):
                            v = j
                            break
                if(v != -1):
                    vis[xx] = True
                    paired[u][v] = paired[v][u] = 1
                    for_ans[xx] = [u, v]
                    adj[u].append(v)
                    adj[v].append(u)
                    dsu.union(u, v)
                    break
                else:
                    continue
    comps = 0
    # print(vis)
    # print(for_ans)
    vis = [0] * n
    helper = []
    # print(adj)
    for i in range(n):
        if(vis[i]):
            continue
        helper.append(i)
        q = deque()
        q.append(i)
        vis[i] = 1
        comps += 1
        if(comps > 2):
            # print(-1)
            # print(n)
            # print(*a)
            # print(-1)
            # assert 1 == 0
            # print("No\n")
            return []
        while q:
            k = len(q)
            for j in range(k):
                node = q.popleft()
                for nbr in adj[node]:
                    if(not vis[nbr]):
                        vis[nbr] = 1
                        q.append(nbr)
    if(comps != 2):
        # assert 1 == 0
        # print("No\n")
        return []
    for_ans[1] = [helper[0], helper[1]]
    return for_ans

def yoyo():
    dsu = DSU(n + 5)
    paired = [[0 for i in range(n)] for i in range(n)]
    cnt = [0] * n
    for_x = [[] for i in range(n)]
    xer = [0] * n
    adj = [[] for i in range(n)]
    vis = [False] * n
    for_ans = [[] for i in range(n)]
    for i in range(2, n):
        diff = [[] for ii in range(i)]
        for j in range(n):
            rem = a[j] % i
            diff[rem].append(j)
        for rem in range(i):
            if(len(diff[rem]) <= 1):
                continue
            for val in diff[rem]:
                cnt[val] += 1
                for_x[val].append(i)
            xer[i] += len(diff[rem])
        if(xer[i] == 2):
            cc = []
            for rem in range(i):
                if(len(diff[rem]) <= 1):
                    continue
                for val in diff[rem]:
                    cnt[val] += 1
                    for_x[val].append(i)
                    cc.append(val)
            if(dsu.find(cc[0]) != dsu.find(cc[1])):
                dsu.union(cc[0], cc[1])
                adj[cc[0]].append(cc[1])
                adj[cc[1]].append(cc[0])
                vis[i] = True
                for_ans[i] = [cc[0], cc[1]]
    mn = [10 ** 8] * n
    l = []
    for i in range(n):
        l.append([cnt[i], i])
    l.sort()
    # print(l)
    for iiiii in range(2):
        for i in range(n):
            c, idx = l[i]
            if(c == 0):
                continue
            for_x[idx]
            nfor_x = []
            for xx in for_x[idx]:
                nfor_x.append([xer[xx], xx])
            nfor_x.sort()
            once = 0
            for tmp, xx in nfor_x:
                if(vis[xx]):
                    continue
                if(iiiii == 0 and once == 0):
                    once = 1
                    continue
                once = 1
                u = idx
                v = -1
                for j in range(n):
                    if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
                        if(a[j] % xx == a[idx] % xx):
                            v = j
                            break
                if(v != -1):
                    vis[xx] = True
                    paired[u][v] = paired[v][u] = 1
                    for_ans[xx] = [u, v]
                    adj[u].append(v)
                    adj[v].append(u)
                    dsu.union(u, v)
                    break
                else:
                    continue
    comps = 0
    # print(vis)
    vis = [0] * n
    helper = []
    # print(adj)
    for i in range(n):
        if(vis[i]):
            continue
        helper.append(i)
        q = deque()
        q.append(i)
        vis[i] = 1
        comps += 1
        if(comps > 2):
            # print(-1)
            # print(n)
            # print(*a)
            # print(-1)
            # assert 1 == 0
            # print("No\n")
            return []
        while q:
            k = len(q)
            for j in range(k):
                node = q.popleft()
                for nbr in adj[node]:
                    if(not vis[nbr]):
                        vis[nbr] = 1
                        q.append(nbr)
    if(comps != 2):
        # assert 1 == 0
        # print("No\n")
        return []
    for_ans[1] = [helper[0], helper[1]]
    return for_ans

def yoyoyo():
    dsu = DSU(n + 5)
    paired = [[0 for i in range(n)] for i in range(n)]
    cnt = [0] * n
    for_x = [[] for i in range(n)]
    xer = [0] * n
    adj = [[] for i in range(n)]
    vis = [False] * n
    for_ans = [[] for i in range(n)]
    for i in range(2, n):
        diff = [[] for ii in range(i)]
        for j in range(n):
            rem = a[j] % i
            diff[rem].append(j)
        for rem in range(i):
            if(len(diff[rem]) <= 1):
                continue
            for val in diff[rem]:
                cnt[val] += 1
                for_x[val].append(i)
            xer[i] += len(diff[rem])
        if(xer[i] == 2):
            cc = []
            for rem in range(i):
                if(len(diff[rem]) <= 1):
                    continue
                for val in diff[rem]:
                    cnt[val] += 1
                    for_x[val].append(i)
                    cc.append(val)
            if(dsu.find(cc[0]) != dsu.find(cc[1])):
                dsu.union(cc[0], cc[1])
                adj[cc[0]].append(cc[1])
                adj[cc[1]].append(cc[0])
                vis[i] = True
                for_ans[i] = [cc[0], cc[1]]
    mn = [10 ** 8] * n
    l = []
    for i in range(n):
        l.append([cnt[i], i])
    l.sort()
    # print(l)
    for iiiii in range(2):
        for i in range(n):
            c, idx = l[i]
            if(c == 0):
                continue
            for_x[idx]
            nfor_x = []
            for xx in for_x[idx]:
                nfor_x.append([xer[xx], xx])
            nfor_x.sort()
            for tmp, xx in nfor_x:
                if(vis[xx]):
                    continue
                if(iiiii < 1):
                    if(randint(0, 1000) % 2 == 0):
                        continue
                u = idx
                v = -1
                for j in range(n):
                    if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
                        if(a[j] % xx == a[idx] % xx):
                            v = j
                            break
                if(v != -1):
                    vis[xx] = True
                    paired[u][v] = paired[v][u] = 1
                    for_ans[xx] = [u, v]
                    adj[u].append(v)
                    adj[v].append(u)
                    dsu.union(u, v)
                    break
                else:
                    continue
    comps = 0
    # print(vis)
    vis = [0] * n
    helper = []
    # print(adj)
    for i in range(n):
        if(vis[i]):
            continue
        helper.append(i)
        q = deque()
        q.append(i)
        vis[i] = 1
        comps += 1
        if(comps > 2):
            # print(-1)
            # print(n)
            # print(*a)
            # print(-1)
            # assert 1 == 0
            # print("No\n")
            return []
        while q:
            k = len(q)
            for j in range(k):
                node = q.popleft()
                for nbr in adj[node]:
                    if(not vis[nbr]):
                        vis[nbr] = 1
                        q.append(nbr)
    if(comps != 2):
        # assert 1 == 0
        # print("No\n")
        return []
    for_ans[1] = [helper[0], helper[1]]
    return for_ans

# Solver function
def solve():
    if(n == 1):
        print("Yes\n")
        return
    # dsu = DSU(n + 5)
    # paired = [[0 for i in range(n)] for i in range(n)]
    # cnt = [0] * n
    # for_x = [[] for i in range(n)]
    # xer = [0] * n
    # adj = [[] for i in range(n)]
    # vis = [False] * n
    # for i in range(2, n):
    #     diff = [[] for ii in range(i)]
    #     for j in range(n):
    #         rem = a[j] % i
    #         diff[rem].append(j)
    #     for rem in range(i):
    #         if(len(diff[rem]) <= 1):
    #             continue
    #         for val in diff[rem]:
    #             cnt[val] += 1
    #             for_x[val].append(i)
    #         xer[i] += len(diff[rem])
    #     if(xer[i] == 2):
    #         cc = []
    #         for rem in range(i):
    #             if(len(diff[rem]) <= 1):
    #                 continue
    #             for val in diff[rem]:
    #                 cnt[val] += 1
    #                 for_x[val].append(i)
    #                 cc.append(val)
    #         if(dsu.find(cc[0]) != dsu.find(cc[1])):
    #             dsu.union(cc[0], cc[1])
    #             adj[cc[0]].append(cc[1])
    #             adj[cc[1]].append(cc[0])
    #             vis[i] = True
    # mn = [10 ** 8] * n
    # l = []
    # for i in range(n):
    #     l.append([cnt[i], i])
    # l.sort()
    # for_ans = [[] for i in range(n)]
    # # print(l)
    # for iiiii in range(2):
    #     for i in range(n):
    #         c, idx = l[i]
    #         if(c == 0):
    #             continue
    #         for_x[idx]
    #         nfor_x = []
    #         for xx in for_x[idx]:
    #             nfor_x.append([xer[xx], xx])
    #         nfor_x.sort()
    #         for tmp, xx in nfor_x:
    #             if(vis[xx]):
    #                 continue
    #             u = idx
    #             v = -1
    #             for j in range(n):
    #                 if(idx != j and not paired[idx][j] and dsu.find(idx) != dsu.find(j)):
    #                     if(a[j] % xx == a[idx] % xx):
    #                         v = j
    #                         break
    #             if(v != -1):
    #                 vis[xx] = True
    #                 paired[u][v] = paired[v][u] = 1
    #                 for_ans[xx] = [u, v]
    #                 adj[u].append(v)
    #                 adj[v].append(u)
    #                 dsu.union(u, v)
    #                 break
    #             else:
    #                 continue
    # comps = 0
    # # print(vis)
    # vis = [0] * n
    # helper = []
    # # print(adj)
    # for i in range(n):
    #     if(vis[i]):
    #         continue
    #     helper.append(i)
    #     q = deque()
    #     q.append(i)
    #     vis[i] = 1
    #     comps += 1
    #     if(comps > 2):
    #         # print(-1)
    #         # print(n)
    #         # print(*a)
    #         # print(-1)
    #         # assert 1 == 0
    #         print("No\n")
    #         return
    #     while q:
    #         k = len(q)
    #         for j in range(k):
    #             node = q.popleft()
    #             for nbr in adj[node]:
    #                 if(not vis[nbr]):
    #                     vis[nbr] = 1
    #                     q.append(nbr)
    # if(comps != 2):
    #     # assert 1 == 0
    #     print("No\n")
    #     return
    # for_ans[1] = [helper[0], helper[1]]
    for_ans = yo()
    if(len(for_ans) == 0):
        for_ans = yoyo()
    if(len(for_ans) == 0):
        for_ans = yoyoyo()
    if(len(for_ans) == 0):
        # assert 1 == 0
        print("No\n")
        return
    print("Yes\n")
    for i in range(1, n):
        u, v = for_ans[i]
        print(str(u + 1) + " " + str(v + 1) + "\n")
        assert (abs(a[u] - a[v]) % i) == 0

# Main
for _ in range(int(input())):
    n = int(input())
    a = [int(x) for x in input().split()]
    solve()

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

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

Think step by step.

Response:
To generate a test input that exposes bugs in the given code, we need to focus on edge cases and scenarios that can challenge the logic and edge conditions of the algorithm. Here are the steps to create such a test input:

1. **Single Vertex Case (n = 1)**:
   - This case is straightforward and should always output "Yes" with no operations required.
   - Input: `1\n1`

2. **Two Vertices Case (n = 2)**:
   - This case is also simple but ensures that the code can handle the smallest non-trivial graph.
   - Input: `2\n1 4`

3. **Three Vertices Case (n = 3)**:
   - This case should test the basic logic of connecting vertices with the smallest possible operations.
   - Input: `3\n1 2 3`

4. **Vertices with Identical Values (n = 3)**:
   - This case tests the ability to handle identical values, which should always be connectable.
   - Input: `3\n1 1 1`

5. **Vertices with Large Values (n = 3)**:
   - This case tests the ability to handle large values and ensure the modulo operations are correct.
   - Input: `3\n1000000000 1000000001 1000000002`

6. **Vertices with Values that are Not Easily Divisible (n = 4)**:
   - This case tests the ability to handle values that are not easily divisible by small numbers.
   - Input: `4\n1 3 5 7`

7. **Vertices with Values that Have Limited Divisibility (n = 4)**:
   - This case tests the ability to handle values that have limited divisibility options.
   - Input: `4\n2 4 6 8`

8. **Vertices with Values that Require Multiple Operations (n = 5)**:
   - This case tests the ability to handle a larger number of vertices and ensure the operations are performed correctly.
   - Input: `5\n1 3 5 7 9`

9. **Vertices with Values that Form a Complex Graph (n = 5)**:
   - This case tests the ability to handle more complex graph structures.
   - Input: `5\n10 20 30 40 50`

10. **Vertices with Values that Form a Disconnected Graph (n = 5)**:
    - This case tests the ability to handle a disconnected graph and ensure the output is "No".
    - Input: `5\n1 2 3 4 5`

Here is the final test input combining these cases:

```plaintext
10
1
1
2
1 4
3
1 2 3
3
1 1 1
3
1000000000 1000000001 1000000002
4
1 3 5 7
4
2 4 6 8
5
1 3 5 7 9
5
10 20 30 40 50
5
1 2 3 4 5
```

This test input covers a range of edge cases and should help identify any bugs in the given code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
#include<iostream>

using namespace std;

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

    int t;
    cin >> t;
    for (int i = 0; i<t;i++){
        int n;
        cin >> n;
        if ((n%4) == 0){
            cout<< n/4 << "\n";
        }else{
            cout << (n/4) +1<< "\n";

        }
    }

    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
3
4
5
6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
#include<iostream>
using namespace std;

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

        if(n <= 3)
        {
            cout << n << "\n";
        }
        else{

            long long result = 1;
            while (n > 3)
            {
                n /= 4;
                result *= 2;
            }
            cout << result << "\n";
        }
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
4
1
5
13
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
#include <iostream>

using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        if (n % 4 != 0)
            cout << n / 4 + 1 << endl;
        else
            cout << n / 4 << 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
3
4
13
1024
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
ran = int(input())
for i in range(ran):
    n = int(input())
    if (n<=4):
        print(1)
    else:
        c = 1
        while n>=4:
            n=n//4
            c*=2
        print(c)

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

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

Think step by step.

Response:
```plaintext
4
1
5
16
65
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
from math import *

t=int(input())
for _ in range(t):
    n=int(input())
    if n<=3:
        print(n)
    else:
        d=1
        while n>1:
            n//=4
            d*=2
        while n<1:
            d//=2
            n+=1
        print(d)

**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
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
import math
for i in range(int(input())):
    x = int(input())
    if x < 4 :
        print(x)
    else:
        max_coins = 1
        while x > 3:
            max_coins *= 2
            x //= 4
        print(max_coins)

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

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

Think step by step.

Response:
```plaintext
4
1
5
17
1000000000000000001
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

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

public class Main {
    static int mod=(int)(1e9+7);
    public static void main(String[] args) {
        int T=Int();
        while(T-->0){
            long x=Lon();
            int cnt=0;
            long ans=1;
            while(x>=4){
                x/=4;
                cnt++;
            }
            pw.println(ans<<cnt);
        }
        pw.flush();
    }
    public static long gcd(long a,long b){
        return b==0?a:gcd(b,a%b);
    }
    public static long lcm(long a,long b){
        return a*b/gcd(a,b);
    }
    private static BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
    private static PrintWriter pw=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    private static StreamTokenizer st=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    public static  int Int(){
        try{
            st.nextToken();
        }catch(IOException e){
            e.printStackTrace();
        }
        return (int) st.nval;
    }
    public static  double Dou(){
        try{
            st.nextToken();
        }catch(IOException e){
            e.printStackTrace();
        }
        return st.nval;
    }
    public static  long Lon(){
        try{
            st.nextToken();
        }catch(IOException e){
            e.printStackTrace();
        }
        return (long) st.nval;
    }
    public static String Str(){
        String p="";
        try{
            p=bf.readLine();
        }catch (IOException e){
            e.printStackTrace();
        }
        return p;
    }

}

**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 identify a challenging test input that exposes the bug in the given buggy code, we need to understand the logic and potential issues in the code. The key points to consider are:

1. **Input Handling**: The code reads input using `StreamTokenizer` and `BufferedReader`.
2. **Operation Logic**: The code performs the transformation `x /= 4` and increments a counter `cnt` each time the operation is performed.
3. **Output Calculation**: The code calculates the result as `ans << cnt`, where `ans` is initialized to 1.

The main issue in the code is that it does not handle the case where `x` is less than 4 correctly. Specifically, the code does not account for the fact that the transformation operation can only be performed if `x` is greater than 3. If `x` is 3 or less, the code should not perform the transformation.

To expose this bug, we can use an input where `x` is 3. This should result in the output being 1, as no transformation can be performed. However, the buggy code might incorrectly handle this case.

Here is the test input:

```plaintext
1
3
```

This input will help identify if the code correctly handles the case where `x` is 3. If the code is correct, it should output 1. If the code is buggy, it might produce an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
import java.util.*;
public class CoinTransform{
  public static void main(String []args)
  {
    Scanner sc=new Scanner(System.in);
    int t=sc.nextInt();
    while(t-->0)
    {
      long n=sc.nextLong();
      if(n<3)
      {
        System.out.println("1");
      }
      else{
      int ans=0;int i=1;
      while(n>3)
      {
        n=n/4;
        ans=(int)Math.pow(2,i++);
      }
      System.out.println(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
4
1
5
13
100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
void solve()
{
	long long int n;
	cin>>n;
	long long int p = (log2(n))/2;
	long long int q = pow(2,p);
	cout<<q<<endl;
}
int main()
{
	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
4
1
5
16
13
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define int long long
#define francesco_bernoulli  ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define yes {cout << "YES\n";}
#define no {cout << "NO\n";}
#define sortx(X) sort(X.begin(),X.end());
#define allr(x) (x).rbegin(),(x).rend()
#define all(x) (x).begin(), (x).end()
#define fx(n) fixed<<setprecision(n)
#define mine(v) *min_element(all(v))
#define maxe(v) *max_element(all(v))
/*---------------------------------------------------------------------------------------------------------------------------------------*/
bool temp(int a,int b){
if(a<=b){
    return true;
}
}
void VoidVortexOfNothingness(){
    int x;
    cin>>x;
    if(x<4){
        cout<<1<<endl;
        return;
    }
    int y=(log(x)/log(4)),z=1;
    for(int i=0;i<y;i++){
        z*=2;
    }
    cout<<z<<endl;
}
signed main()
{
    //freopen("mex.in", "r", stdin);
    //freopen("window.in", "r", stdin);
    //freopen("output.txt", "w", stdout);
   francesco_bernoulli
   int t=1;
    cin>>t;
   while(t--){
   VoidVortexOfNothingness();
   }

    return 0;
}
//CIAO_McQUEEN//

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

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

Think step by step.

Response:
```plaintext
4
1
4
5
6
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
import math
t = int(input())
for test in range(t):
    n = int(input())
    ans = 1
    if n <= 3:
        ans = 1
    else:
        while n > 3:
            ans *= 2
            n = math.floor(n/4)
    print(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
1
4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Coin Transformation
time_limit_ms: 2000
memory_limit_mb: 512
Description: Initially, you have a coin with value $$$n$$$. You can perform the following operation any number of times (possibly zero):

- transform one coin with value $$$x$$$, where $$$x$$$ is greater than $$$3$$$ ($$$x>3$$$), into two coins with value $$$\lfloor \frac{x}{4} \rfloor$$$.

What is the maximum number of coins you can have after performing this operation any number of times?

Input: The first line contains one integer $$$t$$$ ($$$1 \le t \le 10^4$$$) — the number of test cases.

Each test case consists of one line containing one integer $$$n$$$ ($$$1 \le n \le 10^{18}$$$).

Output: For each test case, print one integer — the maximum number of coins you can have after performing the operation any number of times.

Examples:
input:
4
1
5
16
1000000000000000000
output:
1
2
4
536870912

Note: In the first example, you have a coin of value $$$1$$$, and you can't do anything with it. So, the answer is $$$1$$$.

In the second example, you can transform a coin of value $$$5$$$ into two coins with value $$$1$$$.

In the third example, you can transform a coin of value $$$16$$$ into two coins with value $$$4$$$. Each of the resulting coins can be transformed into two coins with value $$$1$$$.

Buggy Code:
n = int(input())
a =1
a1 = 0
for i in range(n):
    m = int(input())
    if m <=3:
        print('1')
    elif m >= 4:
        a1 = 1
        while m >= 4:
            m = m/4
            a1 = a1*2
        print(a1)

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

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

Think step by step.

Response:
```plaintext
5
3
4
5
6
7
```

### Explanation:
1. **n = 3**: This is a boundary case where the coin value is exactly 3, and no transformation can be performed. The expected output is 1.
2. **n = 4**: This is the smallest value where a transformation can be performed, resulting in two coins with value 1. The expected output is 2.
3. **n = 5**: This value can be transformed into two coins with value 1. The expected output is 2.
4. **n = 6**: This value can be transformed into two coins with value 1. The expected output is 2.
5. **n = 7**: This value can be transformed into two coins with value 1. The expected output is 2.

These test cases cover the boundary conditions and small values where the transformation can be easily visualized. They are designed to expose any issues with the logic in the buggy code, particularly how it handles the transformation and the loop conditions.