**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Easy Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the easy version of the problem. The only difference between the two versions are the allowed characters in $s$. In the easy version, $s$ only contains the character ?. You can make hacks only if both versions of the problem are solved.

You are given a permutation $p$ of length $n$. You are also given a string $s$ of length $n$, consisting only of the character ?.

For each $i$ from $1$ to $n$:

  * Define $l_i$ as the largest index $j < i$ such that $p_j > p_i$. If there is no such index, $l_i := i$.    * Define $r_i$ as the smallest index $j > i$ such that $p_j > p_i$. If there is no such index, $r_i := i$.

Initially, you have an undirected graph with $n$ vertices (numbered from $1$ to $n$) and no edges. Then, for each $i$ from $1$ to $n$, add one edge to the graph:

  * If $s_i =$ L, add the edge $(i, l_i)$ to the graph.    * If $s_i =$ R, add the edge $(i, r_i)$ to the graph.    * If $s_i =$ ?, either add the edge $(i, l_i)$ or the edge $(i, r_i)$ to the graph at your choice.

Find the maximum possible diameter$^{\text{∗}}$ over all connected graphs that you can form. Output $-1$ if it is not possible to form any connected graphs.

$^{\text{∗}}$ Let $d(s, t)$ denote the smallest number of edges on any path between $s$ and $t$.

The diameter of the graph is defined as the maximum value of $d(s, t)$ over all pairs of vertices $s$ and $t$.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 4 \cdot 10^5$) — the length of the permutation $p$.

The second line of each test case contains $n$ integers $p_1,p_2,\ldots, p_n$ ($1 \le p_i \le n$) — the elements of $p$, which are guaranteed to form a permutation.

The third line of each test case contains a string $s$ of length $n$. It is guaranteed that it consists only of the character ?.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output the maximum possible diameter over all connected graphs that you form, or $-1$ if it is not possible to form any connected graphs.

Examples:
input:
8
5
2 1 4 3 5
?????
2
1 2
??
3
3 1 2
???
7
5 3 1 6 4 2 7
???????
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
4
1
2
6
4
5
5
8

Note: In the first test case, here are some possible connected graphs that you can form (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)| ![](CDN_BASE_URL/bb06c0111d994ef8c2f7e8e1cb93e921)   ---|---|---      In the second test case, the only connected graph has a diameter of $1$.

Buggy Code:
#include <bits/stdc++.h>
typedef long long i64;
#define F(x, y, z) for (int x = (y); x <= (z); ++x)
#define DF(x, y, z) for (int x = (y); x >= (z); --x)
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define pii pair<int, int>
#define mp make_pair
#define mem(x, y) memset(x, y, sizeof(x))
#define vi vector<int>
#define vi64 vector<i64>
using namespace std;
int read() {
    char ch = getchar();
    int x = 0, pd = 0;
    while (ch < '0' || ch > '9') pd |= ch == '-', ch = getchar();
    while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
    return pd ? -x : x;
}
const int maxn = 400005;
int n, a[maxn], b[maxn], len, lp[maxn], rp[maxn];
char s[maxn];
vi to[maxn];
int F[maxn];
int dep[maxn], id[maxn], rk[maxn], tot, lid[maxn], rid[maxn];
void dfs1(int u) {
    rk[id[u] = ++tot] = u;
    lid[u] = id[u];
    for (auto v : to[u]) dep[v] = dep[u] + 1, dfs1(v);
    rid[u] = tot;
}
#define ls (o << 1)
#define rs (o << 1 | 1)
#define lson ls,l,mid
#define rson rs,mid+1,r
int mx[maxn << 2];
int _max(int x, int y) { return dep[x] > dep[y] ? x : y; }
void pp(int o) { mx[o] = _max(mx[ls], mx[rs]); }
void build(int o, int l, int r) {
    if (l == r) {
        mx[o] = rk[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(lson), build(rson), pp(o);
}
int qry(int o, int l, int r, int L, int R) {
    if (L > R) return -1;
    if (L <= l && r <= R) return mx[o];
    int mid = (l + r) >> 1;
    if (R <= mid) return qry(lson, L, R);
    if (L > mid) return qry(rson, L, R);
    return _max(qry(lson, L, R), qry(rson, L, R));
}
int ans, mxd[maxn];
void solve(int u) {
    int mx1 = 0, mx2 = 0;
    mxd[u] = 0;
    for (auto v : to[u]) {
        solve(v), mxd[u] = max(mxd[u], mxd[v] + 1);
        if (mxd[v] + 1 > mx1) mx2 = mx1, mx1 = mxd[v] + 1;
        else mx2 = max(mx2, mxd[v] + 1);
    }
    ans = max(ans, mx1 + mx2);

    if (F[u]) {
        // cout << a[u] << ' ' << a[F[u]] << endl;
        ans = max(ans, mx1 + 1 + _max(qry(1, 1, n, lid[F[u]], lid[u] - 1), qry(1, 1, n, rid[u] + 1, rid[F[u]]))[dep] - dep[F[u]]);
    }
}
void work() {
    n = read();
    F(i, 1, n) a[i] = read(), to[i].clear(), F[i] = 0;
    scanf("%s", s + 1);
    len = 0;
    int root = 0;
    F(i, 1, n) {
        if (a[i] == n) root = i;
        while (len && a[b[len]] < a[i]) --len;
        lp[i] = b[len];
        b[++len] = i;
    }
    len = 0;
    DF(i, n, 1) {
        while (len && a[b[len]] < a[i]) --len;
        rp[i] = b[len];
        b[++len] = i;
    }
    F(i, 1, n) {
        if (s[i] == 'L') {
            if (!lp[i]) {
                puts("-1");
                return;
            }
            to[lp[i]].eb(i);
        } else if (s[i] == 'R') {
            if (!rp[i]) {
                puts("-1");
                return;
            }
            to[rp[i]].eb(i);
        } else {
            if (i == root) continue;
            if ((a[lp[i]] > a[rp[i]] && rp[i]) || !lp[i]) {
                to[rp[i]].eb(i);
                F[i] = lp[i];
            } else {
                to[lp[i]].eb(i);
                F[i] = rp[i];
            }
        }
    }
    tot = 0, dep[root] = 0, dfs1(root);
    build(1, 1, n);
    ans = 0, solve(root);
    printf("%d\n", ans);
}
int main() {
    int T = read();
    while (T--) work();
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios that could cause the algorithm to fail. Here are the steps to think through this:

1. **Understanding the Bug**: The code provided has a complex structure, but one common issue in such problems is handling the root of the permutation and the edge cases where the permutation has specific patterns.

2. **Edge Cases**:
   - **Single Element**: This is trivial and not very useful for testing.
   - **Two Elements**: This is also simple but can help verify basic functionality.
   - **Permutation with a clear root and specific edge cases**:
     - **All elements in increasing order**: This should form a simple linear graph.
     - **All elements in decreasing order**: This should also form a simple linear graph.
     - **Permutations with a clear root and mixed order**: This can test the handling of more complex graphs.

3. **Test Input**:
   - We will use a permutation where the root is clearly defined and the elements are in a mixed order to test the handling of the `lp` and `rp` arrays and the construction of the graph.

Given these considerations, let's construct a small-scale test input:

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

### Explanation:
- **Permutation**: `3 1 4 2 5`
  - The root is `5` (the largest element).
  - The elements are in a mixed order, which will test the handling of both `lp` and `rp` arrays.
- **String**: `?????`
  - This allows flexibility in choosing edges, which will test the decision-making logic in the code.

This test input is small enough to be manageable and complete, and it should expose any issues in the handling of the permutation and the construction of the graph.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Easy Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the easy version of the problem. The only difference between the two versions are the allowed characters in $s$. In the easy version, $s$ only contains the character ?. You can make hacks only if both versions of the problem are solved.

You are given a permutation $p$ of length $n$. You are also given a string $s$ of length $n$, consisting only of the character ?.

For each $i$ from $1$ to $n$:

  * Define $l_i$ as the largest index $j < i$ such that $p_j > p_i$. If there is no such index, $l_i := i$.    * Define $r_i$ as the smallest index $j > i$ such that $p_j > p_i$. If there is no such index, $r_i := i$.

Initially, you have an undirected graph with $n$ vertices (numbered from $1$ to $n$) and no edges. Then, for each $i$ from $1$ to $n$, add one edge to the graph:

  * If $s_i =$ L, add the edge $(i, l_i)$ to the graph.    * If $s_i =$ R, add the edge $(i, r_i)$ to the graph.    * If $s_i =$ ?, either add the edge $(i, l_i)$ or the edge $(i, r_i)$ to the graph at your choice.

Find the maximum possible diameter$^{\text{∗}}$ over all connected graphs that you can form. Output $-1$ if it is not possible to form any connected graphs.

$^{\text{∗}}$ Let $d(s, t)$ denote the smallest number of edges on any path between $s$ and $t$.

The diameter of the graph is defined as the maximum value of $d(s, t)$ over all pairs of vertices $s$ and $t$.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 4 \cdot 10^5$) — the length of the permutation $p$.

The second line of each test case contains $n$ integers $p_1,p_2,\ldots, p_n$ ($1 \le p_i \le n$) — the elements of $p$, which are guaranteed to form a permutation.

The third line of each test case contains a string $s$ of length $n$. It is guaranteed that it consists only of the character ?.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output the maximum possible diameter over all connected graphs that you form, or $-1$ if it is not possible to form any connected graphs.

Examples:
input:
8
5
2 1 4 3 5
?????
2
1 2
??
3
3 1 2
???
7
5 3 1 6 4 2 7
???????
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
4
1
2
6
4
5
5
8

Note: In the first test case, here are some possible connected graphs that you can form (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)| ![](CDN_BASE_URL/bb06c0111d994ef8c2f7e8e1cb93e921)   ---|---|---      In the second test case, the only connected graph has a diameter of $1$.

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

#define rep(i, l, r) for (int i = (l); i <= (r); ++i)
#define per(i, l, r) for (int i = (r); i >= (l); --i)

using namespace std;

using ll = long long;
using db = long double;
using str = string;

#define mp make_pair
#define fi first
#define se second

#define ttt template <typename T
ttt > using V = vector<T>;
ttt, size_t N > using A = array<T, N>;

#define sz(x) int((x).size())
#define all(x, l, r) &(x)[l], &(x)[r] + 1
#define pb push_back
#define eb emplace_back

#define il inline
ttt > il bool ckmin(T& x, const T& y) { return y < x && (x = y, true); }
ttt > il bool ckmax(T& x, const T& y) { return x < y && (x = y, true); }

il int p2(int n) { return 1 << n; }
il int msk2(int n) { return p2(n) - 1; }
il int msb(int n) { return n == 0 ? -1 : 31 ^ __builtin_clz(n); }
il int bits(int n) { return 1 + msb(n - 1); }
il int bceil(int n) { return 1 << bits(n); }

constexpr int N = 4e5 + 5, Log = 19;
int n, ts, ti;
A<int, N> p, stk, top, len, dep, dfn, low;
A<A<int, N>, Log> mx;
A<A<int, 2>, N> ch;
A<V<int>, N> ver;
str s;

void dfs(int u, int p) {
  len[u] = 0, dfn[u] = ++ti, ver[top[u]].pb(u);
  rep (i, 0, 1) {
    int v = ch[u][i];
    if (!v) continue;
    top[v] = i == (u == ch[p][1]) ? top[u] : p;
    dep[v] = dep[u] + 1;
    dfs(v, u);
    ckmax(len[u], len[v] + 1);
  }
  low[u] = ti;
}
void init() {
  rep (i, 1, n) mx[0][dfn[i]] = dep[i];
  rep (i, 0, Log)
    rep (j, 1, n - p2(i + 1) + 1)
      mx[i + 1][j] = max(mx[i][j], mx[i][j + p2(i)]);
}
int qry(int l, int r) {
  if (l > r) return 0;
  int k = msb(r - l + 1);
  return max(mx[k][l], mx[k][r - p2(k) + 1]);
}
void proc() {
  cin >> n, ts = 0;
  fill_n(&ch[1][0], 2 * n, 0);
  rep (i, 1, n) {
    cin >> p[i];
    int old = ts;
    while (ts && p[i] > p[stk[ts]]) --ts;
    if (ts) ch[stk[ts]][1] = i;
    if (ts < old) ch[i][0] = stk[ts + 1];
    stk[++ts] = i;
  }
  cin >> s;
  rep (i, 1, n) ver[i].clear();
  int rt = stk[1];
  ti = dep[rt] = top[rt] = 0, dfs(rt, 0);
  init(), len[0] = -1;
  int ans = 0;
  rep (i, 1, n) {
    ckmax(ans, len[ch[i][0]] + len[ch[i][1]] + 2);
    for (int j : ver[i]) {
      int mxDep = max(qry(dfn[i], dfn[j] - 1), qry(low[j] + 1, low[i]));
      ckmax(ans, len[j] + 1 + mxDep - dep[i]);
    }
  }
  cout << ans << "\n";
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false), cin.exceptions(cin.failbit);
  int t;
  cin >> t;
  rep (i, 1, t) proc();
}

**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
4 3 2 1
????
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Easy Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the easy version of the problem. The only difference between the two versions are the allowed characters in $s$. In the easy version, $s$ only contains the character ?. You can make hacks only if both versions of the problem are solved.

You are given a permutation $p$ of length $n$. You are also given a string $s$ of length $n$, consisting only of the character ?.

For each $i$ from $1$ to $n$:

  * Define $l_i$ as the largest index $j < i$ such that $p_j > p_i$. If there is no such index, $l_i := i$.    * Define $r_i$ as the smallest index $j > i$ such that $p_j > p_i$. If there is no such index, $r_i := i$.

Initially, you have an undirected graph with $n$ vertices (numbered from $1$ to $n$) and no edges. Then, for each $i$ from $1$ to $n$, add one edge to the graph:

  * If $s_i =$ L, add the edge $(i, l_i)$ to the graph.    * If $s_i =$ R, add the edge $(i, r_i)$ to the graph.    * If $s_i =$ ?, either add the edge $(i, l_i)$ or the edge $(i, r_i)$ to the graph at your choice.

Find the maximum possible diameter$^{\text{∗}}$ over all connected graphs that you can form. Output $-1$ if it is not possible to form any connected graphs.

$^{\text{∗}}$ Let $d(s, t)$ denote the smallest number of edges on any path between $s$ and $t$.

The diameter of the graph is defined as the maximum value of $d(s, t)$ over all pairs of vertices $s$ and $t$.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 4 \cdot 10^5$) — the length of the permutation $p$.

The second line of each test case contains $n$ integers $p_1,p_2,\ldots, p_n$ ($1 \le p_i \le n$) — the elements of $p$, which are guaranteed to form a permutation.

The third line of each test case contains a string $s$ of length $n$. It is guaranteed that it consists only of the character ?.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output the maximum possible diameter over all connected graphs that you form, or $-1$ if it is not possible to form any connected graphs.

Examples:
input:
8
5
2 1 4 3 5
?????
2
1 2
??
3
3 1 2
???
7
5 3 1 6 4 2 7
???????
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
4
1
2
6
4
5
5
8

Note: In the first test case, here are some possible connected graphs that you can form (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)| ![](CDN_BASE_URL/bb06c0111d994ef8c2f7e8e1cb93e921)   ---|---|---      In the second test case, the only connected graph has a diameter of $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const int N=300100;
typedef long long ll;
const ll mod=998244353;
const int K=200;
const int inf=1e9;
vector<vector<int> >g;
vector<int>tin,tout;
vector<int>d;
int timer=0;
ll ANS=0;
vector<int>h;
void dfs(int v){
  tin[v]=++timer;
  d[v]=0;
  for (int to:g[v]){
    h[to]=h[v]+1;
    dfs(to);
    ANS=max(ANS,d[v]+d[to]+1ll);
    d[v]=max(d[v],d[to]+1);
  }
  tout[v]=timer;
}
struct SegTree{
  vector<ll>t;
  vector<ll>w;
  SegTree(int n){
    t.resize(4*(n+1),0);
    w.resize(4*(n+1),0);
  }
  void push(int v){
    t[v+v]+=w[v];
    t[v+v+1]+=w[v];
    w[v+v]+=w[v];
    w[v+v+1]+=w[v];
    w[v]=0;
  }
  void upd(int v,int l,int r,int L,int R,int x){
    if (l>R || r<L) return;
    if (l>=L && r<=R){
      t[v]+=x;
      w[v]+=x;
      return;
    }
    push(v);
    int m=(l+r)/2;
    upd(v+v,l,m,L,R,x);
    upd(v+v+1,m+1,r,L,R,x);
    t[v]=max(t[v+v],t[v+v+1]);
  }
  ll get(int v,int l,int r,int L,int R){
    if (l>R || r<L) return 0;
    if (l>=L && r<=R) return t[v];
    push(v);
    int m=(l+r)/2;
    return max(get(v+v,l,m,L,R),get(v+v+1,m+1,r,L,R));
  }

};
void solve(){
  g.clear();
  tin.clear();
  tout.clear();
  d.clear();
  h.clear();
  ANS=0;
  timer=0;
  int n;cin>>n;
  g.resize(n+1);
  tin.resize(n+1);
  tout.resize(n+1);
  d.resize(n+1);
  h.resize(n+1);
  vector<int>p(n+2,0);
  for (int i=1;i<=n;i++) cin>>p[i];
  string s;cin>>s;
  p[0]=p[n+1]=n+1;
  vector<int>l(n+1,0);
  for (int i=1;i<=n;i++){
    l[i]=0;
    for (int j=i-1;j>=1;j=l[j]){
      if (p[j]>p[i]){
        l[i]=j;
        break;
      }
    }
  }
  vector<int>r(n+1,n+1);
  for (int i=n;i>=1;i--){
    r[i]=n+1;
    for (int j=i+1;j<=n;j=r[j]){
      if (p[j]>p[i]){
        r[i]=j;
        break;
      }
    }
  }
  int root=-1;
  for (int i=1;i<=n;i++){
    if (p[i]==n){
      root=i;
      continue;
    }
    if (p[l[i]]>p[r[i]]){
      swap(l[i],r[i]);
    }
    g[l[i]].push_back(i);
    // cout<<l[i]<<" "<<i<<endl;
  }
  h[root]=0;
  dfs(root);
  SegTree T(n);
  for (int i=1;i<=n;i++){
    if (tin[i]<tout[i]){
      T.upd(1,1,n,tin[i]+1,tout[i],1);
    }
  }

  for (int i=1;i<=n;i++){
    if (p[r[i]]==n+1) continue;
    T.upd(1,1,n,tin[i],tout[i],-inf);
    ANS=max(ANS,d[i]+1+T.get(1,1,n,tin[r[i]],tout[r[i]])-h[r[i]]);
    T.upd(1,1,n,tin[i],tout[i],inf);
  }
  cout<<ANS<<endl;

}
int32_t main(){
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt=1;
  cin>>tt;
  while (tt--){
    solve();
  }
  // cerr<<clock()/(double)CLOCKS_PER_SEC<<endl;
  return 0;
}

/*
9 8
bb-bbb-ba
*a-a-*-*

*/

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Easy Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the easy version of the problem. The only difference between the two versions are the allowed characters in $s$. In the easy version, $s$ only contains the character ?. You can make hacks only if both versions of the problem are solved.

You are given a permutation $p$ of length $n$. You are also given a string $s$ of length $n$, consisting only of the character ?.

For each $i$ from $1$ to $n$:

  * Define $l_i$ as the largest index $j < i$ such that $p_j > p_i$. If there is no such index, $l_i := i$.    * Define $r_i$ as the smallest index $j > i$ such that $p_j > p_i$. If there is no such index, $r_i := i$.

Initially, you have an undirected graph with $n$ vertices (numbered from $1$ to $n$) and no edges. Then, for each $i$ from $1$ to $n$, add one edge to the graph:

  * If $s_i =$ L, add the edge $(i, l_i)$ to the graph.    * If $s_i =$ R, add the edge $(i, r_i)$ to the graph.    * If $s_i =$ ?, either add the edge $(i, l_i)$ or the edge $(i, r_i)$ to the graph at your choice.

Find the maximum possible diameter$^{\text{∗}}$ over all connected graphs that you can form. Output $-1$ if it is not possible to form any connected graphs.

$^{\text{∗}}$ Let $d(s, t)$ denote the smallest number of edges on any path between $s$ and $t$.

The diameter of the graph is defined as the maximum value of $d(s, t)$ over all pairs of vertices $s$ and $t$.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 4 \cdot 10^5$) — the length of the permutation $p$.

The second line of each test case contains $n$ integers $p_1,p_2,\ldots, p_n$ ($1 \le p_i \le n$) — the elements of $p$, which are guaranteed to form a permutation.

The third line of each test case contains a string $s$ of length $n$. It is guaranteed that it consists only of the character ?.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output the maximum possible diameter over all connected graphs that you form, or $-1$ if it is not possible to form any connected graphs.

Examples:
input:
8
5
2 1 4 3 5
?????
2
1 2
??
3
3 1 2
???
7
5 3 1 6 4 2 7
???????
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
4
1
2
6
4
5
5
8

Note: In the first test case, here are some possible connected graphs that you can form (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)| ![](CDN_BASE_URL/bb06c0111d994ef8c2f7e8e1cb93e921)   ---|---|---      In the second test case, the only connected graph has a diameter of $1$.

Buggy Code:
//#pragma GCC optimize("Ofast","inline","-ffast-math")
//#pragma GCC target("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
//#define int long long
using namespace __gnu_pbds;
using namespace std;

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

int n,p[400010],st[400010][20],rev[400010],L[400010],R[400010],dep[400010],ans;
char ch[400010];

inline int maxv(int l,int r)
{
	int id=rev[r-l+1];
	if(p[st[l][id]]>p[st[r-(1<<id)+1][id]]) return st[l][id];
	else return st[r-(1<<id)+1][id];
}

inline int dfs(int l,int r)
{
	int pos=maxv(l,r);
	if(l<pos) L[pos]=dfs(l,pos-1);
	else L[pos]=0;
	if(pos<r) R[pos]=dfs(pos+1,r);
	else R[pos]=0;
	dep[pos]=0;
	if(L[pos]) dep[pos]=max(dep[pos],dep[L[pos]]+1);
	if(R[pos]) dep[pos]=max(dep[pos],dep[R[pos]]+1);
	return pos;
}

inline void calc(int x,int flag,int len,int dis)
{
	if(L[x]){
		if(flag==1) calc(L[x],1,len+1,dis+1);
		else calc(L[x],1,1,dis+1);
	}
	if(R[x]){
		if(flag==2) calc(R[x],2,len+1,dis+1);
		else calc(R[x],2,1,dis+1);
	}
	int ll=0,rr=0;
	if(L[x]) ll=dep[L[x]]+1;
	if(R[x]) rr=dep[R[x]]+1;
	int off=0;
	if(flag==1){
		if(R[x]) off=max(off,1);
		if(L[x]&&len<dis) off=max(off,len+1);
	}
	else if(flag==2){
		if(L[x]) off=max(off,1);
		if(R[x]&&len<dis) off=max(off,len+1);
	}
	ans=max(ans,ll+rr+off);
}

signed main()
{
	ios::sync_with_stdio(false);cin.tie(0);
	int t;cin>>t;
	while(t--){
		cin>>n;
		for(int i=1;i<=n;i++) cin>>p[i];
		for(int i=1;i<=n;i++) cin>>ch[i];
		rev[1]=0;
		for(int i=2;i<=n;i++) rev[i]=rev[i/2]+1;
		for(int i=1;i<=n;i++) st[i][0]=i;
		for(int i=1;i<20;i++){
			for(int j=1;j+(1<<i)-1<=n;j++){
				if(p[st[j][i-1]]>p[st[j+(1<<(i-1))][i-1]]) st[j][i]=st[j][i-1];
				else st[j][i]=st[j+(1<<(i-1))][i-1];
			}
		}
		int root=dfs(1,n);
		ans=0;calc(root,0,0,0);
		cout<<ans<<'\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
3
2
2 1
??
3
3 2 1
???
4
4 3 2 1
????
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Easy Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the easy version of the problem. The only difference between the two versions are the allowed characters in $s$. In the easy version, $s$ only contains the character ?. You can make hacks only if both versions of the problem are solved.

You are given a permutation $p$ of length $n$. You are also given a string $s$ of length $n$, consisting only of the character ?.

For each $i$ from $1$ to $n$:

  * Define $l_i$ as the largest index $j < i$ such that $p_j > p_i$. If there is no such index, $l_i := i$.    * Define $r_i$ as the smallest index $j > i$ such that $p_j > p_i$. If there is no such index, $r_i := i$.

Initially, you have an undirected graph with $n$ vertices (numbered from $1$ to $n$) and no edges. Then, for each $i$ from $1$ to $n$, add one edge to the graph:

  * If $s_i =$ L, add the edge $(i, l_i)$ to the graph.    * If $s_i =$ R, add the edge $(i, r_i)$ to the graph.    * If $s_i =$ ?, either add the edge $(i, l_i)$ or the edge $(i, r_i)$ to the graph at your choice.

Find the maximum possible diameter$^{\text{∗}}$ over all connected graphs that you can form. Output $-1$ if it is not possible to form any connected graphs.

$^{\text{∗}}$ Let $d(s, t)$ denote the smallest number of edges on any path between $s$ and $t$.

The diameter of the graph is defined as the maximum value of $d(s, t)$ over all pairs of vertices $s$ and $t$.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 4 \cdot 10^5$) — the length of the permutation $p$.

The second line of each test case contains $n$ integers $p_1,p_2,\ldots, p_n$ ($1 \le p_i \le n$) — the elements of $p$, which are guaranteed to form a permutation.

The third line of each test case contains a string $s$ of length $n$. It is guaranteed that it consists only of the character ?.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output the maximum possible diameter over all connected graphs that you form, or $-1$ if it is not possible to form any connected graphs.

Examples:
input:
8
5
2 1 4 3 5
?????
2
1 2
??
3
3 1 2
???
7
5 3 1 6 4 2 7
???????
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
4
1
2
6
4
5
5
8

Note: In the first test case, here are some possible connected graphs that you can form (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)| ![](CDN_BASE_URL/bb06c0111d994ef8c2f7e8e1cb93e921)   ---|---|---      In the second test case, the only connected graph has a diameter of $1$.

Buggy Code:
#include <bits/stdc++.h>
#define rep(i,n) for(i=1;i<=(n);i++)
using namespace std;
int t,n,m,i,j,a[400005],ra[400005],ans;
char typ[400005];
int stk[400005],lf[400005],rf[400005],fa[400005],tp;
int dep[400005],lc[400005],rc[400005];
multiset<int> s;
int main()
{
	scanf("%d",&t);
	while(t--){
		ans=-1;scanf("%d",&n);
		rep(i,n)scanf("%d",&a[i]),ra[a[i]]=i;
		rep(i,n){
			while((typ[i]=getchar())!='?'&&!isupper(typ[i]));
			assert(typ[i]=='?');
		}
		tp=0;
		rep(i,n){
			while(tp&&a[stk[tp]]<a[i]){
				rf[stk[tp]]=i;
				tp--;
			}
			stk[++tp]=i;
		}
		while(tp) rf[stk[tp]]=stk[tp],tp--;
		tp=0;
		for(i=n;i;i--){
			while(tp&&a[stk[tp]]<a[i]){
				lf[stk[tp]]=i;
				tp--;
			}
			stk[++tp]=i;
		}
		while(tp) lf[stk[tp]]=stk[tp],tp--;
		rep(i,n){
			if(lf[i]!=i&&(a[lf[i]]<a[rf[i]]||rf[i]==i))fa[i]=lf[i];else fa[i]=rf[i];
		}
		rep(i,n)dep[i]=lc[i]=rc[i]=0;
		rep(i,n-1){
			dep[fa[ra[i]]]=max(dep[fa[ra[i]]],dep[ra[i]]+1);
			if(fa[ra[i]]==lf[ra[i]]) rc[fa[ra[i]]]=ra[i];else lc[fa[ra[i]]]=ra[i];
		}
		rep(i,n) ans=max(ans,dep[lc[i]]+(!!lc[i])+dep[rc[i]]+(!!rc[i]));
		tp=0;
		rep(i,n){
			int cur=-0x3f3f3f3f,cur2=-0x3f3f3f3f;
			while(tp&&a[stk[tp]]<a[i]){
				cur2=max(cur2,cur+(dep[lc[stk[tp]]]+(!!lc[stk[tp]])));
				cur2++;
				cur=max(cur,dep[stk[tp]]+1);
				tp--;
			}
			ans=max(ans,cur2);
			stk[++tp]=i;
		}
		tp=0;
		for(i=n;i;i--){
			int cur=-0x3f3f3f3f,cur2=-0x3f3f3f3f;
			while(tp&&a[stk[tp]]<a[i]){
				cur2=max(cur2,cur+(dep[rc[stk[tp]]]+(!!rc[stk[tp]])));
				cur2++;
				cur=max(cur,dep[stk[tp]]+1);
				tp--;
			}
			ans=max(ans,cur2);
			stk[++tp]=i;
		}
		printf("%d\n",ans);
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Spinning Round (Easy Version)
time_limit_ms: 7000
memory_limit_mb: 1024
Description: This is the easy version of the problem. The only difference between the two versions are the allowed characters in $s$. In the easy version, $s$ only contains the character ?. You can make hacks only if both versions of the problem are solved.

You are given a permutation $p$ of length $n$. You are also given a string $s$ of length $n$, consisting only of the character ?.

For each $i$ from $1$ to $n$:

  * Define $l_i$ as the largest index $j < i$ such that $p_j > p_i$. If there is no such index, $l_i := i$.    * Define $r_i$ as the smallest index $j > i$ such that $p_j > p_i$. If there is no such index, $r_i := i$.

Initially, you have an undirected graph with $n$ vertices (numbered from $1$ to $n$) and no edges. Then, for each $i$ from $1$ to $n$, add one edge to the graph:

  * If $s_i =$ L, add the edge $(i, l_i)$ to the graph.    * If $s_i =$ R, add the edge $(i, r_i)$ to the graph.    * If $s_i =$ ?, either add the edge $(i, l_i)$ or the edge $(i, r_i)$ to the graph at your choice.

Find the maximum possible diameter$^{\text{∗}}$ over all connected graphs that you can form. Output $-1$ if it is not possible to form any connected graphs.

$^{\text{∗}}$ Let $d(s, t)$ denote the smallest number of edges on any path between $s$ and $t$.

The diameter of the graph is defined as the maximum value of $d(s, t)$ over all pairs of vertices $s$ and $t$.

Input: Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1 \le t \le 2 \cdot 10^4$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($2 \le n \le 4 \cdot 10^5$) — the length of the permutation $p$.

The second line of each test case contains $n$ integers $p_1,p_2,\ldots, p_n$ ($1 \le p_i \le n$) — the elements of $p$, which are guaranteed to form a permutation.

The third line of each test case contains a string $s$ of length $n$. It is guaranteed that it consists only of the character ?.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output the maximum possible diameter over all connected graphs that you form, or $-1$ if it is not possible to form any connected graphs.

Examples:
input:
8
5
2 1 4 3 5
?????
2
1 2
??
3
3 1 2
???
7
5 3 1 6 4 2 7
???????
5
5 2 1 3 4
?????
6
6 2 3 4 5 1
??????
8
1 7 5 6 2 8 4 3
????????
12
6 10 7 1 8 5 12 2 11 3 4 9
????????????
output:
4
1
2
6
4
5
5
8

Note: In the first test case, here are some possible connected graphs that you can form (the labels are indices):

![](CDN_BASE_URL/4561c9025a4357785c7aff0bb1bbff9a)| ![](CDN_BASE_URL/a680f9e3bffa38fd90dd7bbc4c3e5f3b)| ![](CDN_BASE_URL/bb06c0111d994ef8c2f7e8e1cb93e921)   ---|---|---      In the second test case, the only connected graph has a diameter of $1$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int p[400005];
char s[400005];
int st[400005];
int ls[400005],rs[400005],fa[400005],dep[400005];
int L[400005],R[400005];
void dfs0(int x){
	L[x]=R[x]=x;
	if(ls[x]){
		dep[ls[x]]=dep[x]+1;
		dfs0(ls[x]);L[x]=L[ls[x]];
	}
	if(rs[x]){
		dep[rs[x]]=dep[x]+1;
		dfs0(rs[x]);R[x]=R[rs[x]];
	}
}
int mx[21][400005],lg[400005];
int qmax(int l,int r){
	assert(l-1<=r);
	if(l>r)return 0;
	int cd=lg[r-l+1];
	return max(mx[cd][l],mx[cd][r-(1<<cd)+1]);
}
int hb[400005],tp[400005];
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		for(int i=1;i<=n;++i)scanf("%d",&p[i]);
		scanf("%s",s+1);
		int tt=0;
		for(int i=1;i<=n;++i)ls[i]=rs[i]=fa[i]=dep[i]=0;
		for(int i=1;i<=n;++i){
			int ll=0;
			while(tt&&p[st[tt]]<p[i])ll=st[tt--];
			ls[i]=ll,fa[ll]=i;
			rs[st[tt]]=i,fa[i]=st[tt];
			st[++tt]=i;
		}
		int rt=st[1];
		dfs0(rt);
		lg[1]=0;
		for(int i=2;i<=n;++i)lg[i]=lg[i>>1]+1;
		for(int i=1;i<=n;++i)mx[0][i]=dep[i];
		for(int i=1;i<=20;++i)for(int j=1;j<=n-(1<<i)+1;++j){
			mx[i][j]=max(mx[i-1][j],mx[i-1][j+(1<<(i-1))]);
		}
		int ans=0;
		for(int x=1;x<=n;++x)if(ls[x]||rs[x]){
			int A=0,B=0;
			if(ls[x])A=qmax(L[ls[x]],R[ls[x]]);
			if(rs[x])B=qmax(L[rs[x]],R[rs[x]]);
			tp[x]=0;
			if(!ls[x])tp[x]=B-dep[x];
			else if(!rs[x])tp[x]=A-dep[x];
			else tp[x]=A+B-2*dep[x];
			ans=max(ans,tp[x]);
		}
		stack<int>ss;
		while(ss.size())ss.pop();
		for(int i=1;i<=n;++i)hb[i]=-fa[i];
		for(int i=1;i<=n;++i){
			while(ss.size()&&p[ss.top()]<p[i])ss.pop();
			if(ss.size())hb[i]+=ss.top();
			ss.emplace(i);
		}
		while(ss.size())ss.pop();
		for(int i=n;i>=1;--i){
			while(ss.size()&&p[ss.top()]<p[i])ss.pop();
			if(ss.size())hb[i]+=ss.top();
			ss.emplace(i);
		}
		for(int x=1;x<=n;++x)if(hb[x]){
			int y=hb[x];
			int lx=qmax(L[x],R[x])-dep[x];
			int ly=max(qmax(L[y],L[x]-1),qmax(R[x]+1,R[y]))-dep[y];
			ans=max(ans,lx+ly+1);
		}
		printf("%d\n",ans);
	}
	return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
#include <bits/stdc++.h>
#define fr(n) for(int i=0;i<n;i++)
typedef long long ll;
using namespace std;
int main() {
    ll t;
    cin >> t;
    while (t--) {
   ll n;
   cin>>n;
   ll arr[n];
   fr(n)cin>>arr[i];
   ll maxi=*max_element(arr,arr+n);
   ll ct=0;
   fr(n){
    if(arr[i]==maxi)ct++;
   }
   if(ct%2!=0)cout<<"YES"<<endl;
   else {
    if(n==ct+1 && n%2!=0)cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
   }
    }
    return 0;
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<int>v(n);
        for(int i=0;i<n;i++){
            cin>>v[i];
        }
        int max=v[0];
        for(int i=0;i<n;i++){
            if(v[i]>max){
                max=v[i];
            }
        }
        int ct=0;
        sort(v.rbegin(),v.rend());
        for(int i=0;i<n;i++){
            if(v[i]==max) ct++;
        }
        if(ct%2!=0){
            cout<<"YES"<<endl;
        }
        else if(ct==n && ct%2==0) cout<<"NO"<<endl;
        else{
            int ctmin=0;
            int x;
            for(int i=0;i<n;i++){
                if(v[i]!=max){
                    x=v[i];
                    break;
                }
            }
            for(int i=0;i<n;i++){
                if(v[i]==x){
                    ctmin++;
                }
            }
            if(ctmin%2==0) cout<<"NO"<<endl;
            else cout<<"YES"<<endl;
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

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

#define optimize() ios_base::sync_with_stdio(0); cin.tie(0)
#define ll long long
#define ff first
#define ss second
#define sz size()
#define pb push_back
#define vinput for(auto& x: v) cin >> x;
#define all(x) x.begin(),x.end()
#define allr(x) x.rbegin(),x.rend()
#define range(i,x,y) for(int i=x;i<y;i++)
#define range_back(i,x,y) for(int i=y-1;i>=x;i--)
#define mem(a,b) memset(a, b, sizeof(a) )

typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<double, double> pdd;
typedef vector<pii> vii;

// Debug..
#ifdef LOCAL
#include "debug2.h"
#else
#define dg(x...)
#endif

void Solve() {
    int n;
    cin >> n;
    vi v(n);
    vinput;

    map<int, int>mp;
    int mx = 0;
    range(i, 0, n){
        mp[v[i]]++;
        mx = max(mx, v[i]);
    }

    if((mp[mx]&1) || (mp.sz > 1 && mp[mx]%2)){
        cout << "YES\n";
    }
    else{
        cout << "NO\n";
    }
}

int main() {
    optimize();

    int t = 1;
    cin >> t;
    for(int i=1; i<=t; i++){
        //cout << "Case " << i << ": ";
        Solve();
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's analyze the logic of the code and identify potential edge cases or scenarios where the code might fail.

### Analysis of the Code

1. **Map Usage**: The code uses a map to count the frequency of each element in the array.
2. **Maximum Element**: The code identifies the maximum element in the array.
3. **Winning Condition**: The code checks if the maximum element appears an odd number of times or if there are multiple distinct elements and the maximum element appears an odd number of times.

### Potential Bugs

- **Odd Frequency of Maximum Element**: The code assumes that if the maximum element appears an odd number of times, Alice has a winning strategy. This might not always be true, especially if the array has multiple elements with the same frequency.
- **Multiple Distinct Elements**: The code checks if there are multiple distinct elements and the maximum element appears an odd number of times. This might not cover all edge cases, especially if the array has a complex distribution of elements.

### Challenging Test Input

To expose the bug, we can create a test input where the maximum element appears an odd number of times, but Alice does not have a winning strategy. A good example is an array where the maximum element appears an odd number of times, but the distribution of other elements makes it impossible for Alice to win.

#### Test Input
```plaintext
1
5
3 3 3 2 2
```

### Explanation
- **Array**: [3, 3, 3, 2, 2]
- **Maximum Element**: 3
- **Frequency of Maximum Element**: 3 (odd)
- **Other Elements**: 2, 2

In this case, the code will output "YES" because the maximum element (3) appears an odd number of times. However, let's analyze the game:

1. **Alice's First Move**: Alice can choose any of the 3s. After her move, the array becomes [0, 3, 3, 2, 2] and `mx` is 3.
2. **Bob's Move**: Bob can choose any of the 3s. After his move, the array becomes [0, 0, 3, 2, 2] and `mx` is 3.
3. **Alice's Second Move**: Alice can choose the remaining 3. After her move, the array becomes [0, 0, 0, 2, 2] and `mx` is 3.
4. **Bob's Second Move**: Bob can choose any of the 2s. After his move, the array becomes [0, 0, 0, 0, 2] and `mx` is 3.
5. **Alice's Third Move**: Alice can choose the remaining 2. After her move, the array becomes [0, 0, 0, 0, 0] and `mx` is 3.
6. **Bob's Third Move**: Bob cannot make a move because all elements are 0.

Thus, Bob wins, but the code incorrectly outputs "YES". This test input exposes the bug in the code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
for j in range(int(input())):
    n = int(input())
    arr = [int(i) for i in input().split()]
    m = arr.count(max(arr))

    if m%2!=0:
        print('YES')
    elif (n-m)%2!=0:
        print('YES')
    else:
        print('NO')

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split()))
    a.sort(reverse=True)
    mx = 0
    moves = 0
    for i in range(n):
        if a[i] >= mx:
            mx = a[i]
            a[i] = 0
            moves += 1
        else:
            break
    if moves % 2 == 1:
        print("YES")
    else:
        print("NO")

**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 scenarios where the sorting and the move counting logic might lead to incorrect results. The key is to create a test case where the order of elements and the values of `mx` and `moves` can lead to an incorrect determination of the winner.

### Step-by-Step Analysis:
1. **Understanding the Bug:**
   - The code sorts the array in descending order before processing.
   - This means that the original order of elements in the array is lost.
   - The bug might arise when the original order of elements is significant in determining the winner.

2. **Creating a Test Case:**
   - We need a test case where the order of elements matters.
   - Specifically, we need a case where the sorted order leads to a different outcome than the original order.

3. **Example Test Case:**
   - Consider an array where the highest value is in the middle, and the other values are lower.
   - This can trick the sorting logic into thinking the game should proceed differently than it would in the original order.

### Test Case:
```plaintext
1
4
2 1 4 3
```

### Explanation:
- **Input:**
  - `t = 1` (one test case)
  - `n = 4` (array size)
  - `a = [2, 1, 4, 3]` (array elements)

- **Expected Behavior:**
  - Alice starts first.
  - Alice can choose `i = 3` (0-indexed, 4 in the array) since `a[3] = 4 >= mx = 0`.
  - After Alice's move: `a = [2, 1, 0, 3]` and `mx = 4`.
  - Bob can choose `i = 3` (0-indexed, 3 in the array) since `a[3] = 3 >= mx = 4`.
  - After Bob's move: `a = [2, 1, 0, 0]` and `mx = 4`.
  - Alice can choose `i = 0` (0-indexed, 2 in the array) since `a[0] = 2 >= mx = 4`.
  - After Alice's move: `a = [0, 1, 0, 0]` and `mx = 4`.
  - Bob can choose `i = 1` (0-indexed, 1 in the array) since `a[1] = 1 >= mx = 4`.
  - After Bob's move: `a = [0, 0, 0, 0]` and `mx = 4`.
  - Alice cannot make a move, so Bob wins.

- **Buggy Code Output:**
  - The buggy code sorts the array to `[4, 3, 2, 1]`.
  - Alice can choose `i = 0` (4 in the array).
  - After Alice's move: `a = [0, 3, 2, 1]` and `mx = 4`.
  - Bob can choose `i = 1` (3 in the array).
  - After Bob's move: `a = [0, 0, 2, 1]` and `mx = 4`.
  - Alice can choose `i = 2` (2 in the array).
  - After Alice's move: `a = [0, 0, 0, 1]` and `mx = 4`.
  - Bob can choose `i = 3` (1 in the array).
  - After Bob's move: `a = [0, 0, 0, 0]` and `mx = 4`.
  - Alice cannot make a move, so Bob wins.
  - The number of moves is 4 (even), so the buggy code outputs "NO".

- **Correct Output:**
  - The correct output should be "NO" because Bob wins in both scenarios.

### Final Test Input:
```plaintext
1
4
2 1 4 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
t = int(input())
from collections import Counter
answers = []
for i in range(t):
    n = int(input())
    a = list(map(int, input().split(" ")))
    b = Counter(a)
    c = b[max(a)]
    if c%2 == 1:
        answers.append("Yes")
    else:
        answers.append("No")

for j in answers:
    print(j)

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.StringTokenizer;

public class SubmissionBait {

	public static void main(String[] args) throws IOException{
		// TODO Auto-generated method stub
		FastScanner s = new FastScanner();
		PrintWriter p = new PrintWriter();
		int t = s.nextInt();
		while(t>0) {

			//Your code Here.
			int n = s.nextInt();
			int[] arr = new int[n];

			for(int i=0;i<n;i++)
				arr[i]=s.nextInt();

			int max = Integer.MIN_VALUE;

			for(int i=0;i<n;i++) {
				max = Math.max(max, arr[i]);
			}

			int cnt = 0;
			for(int i=0;i<n;i++) {
				if(arr[i]==max) {
					cnt++;
				}
			}

			if(cnt%2==0) {
				p.println("NO");
			}else {
				p.println("YES");
			}

			//Your code ends.
			t--;
		}

		p.close();
	}
	static class PrintWriter{
		BufferedWriter bwr = new BufferedWriter(new OutputStreamWriter(System.out));

		public void print(Object object) throws IOException{
			bwr.append(""+object);
			bwr.flush();
		}
		public void println(Object object) throws IOException{
			print(object);
			bwr.append("\n");
			bwr.flush();
		}
		public void close() throws IOException{
			bwr.close();
		}

	}
	static class FastScanner{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch(IOException e) {}
			}
			return st.nextToken();

		}
		 int nextInt() {
			 return Integer.parseInt(next());
		 }
		 long nextLong() {
			 return Long.parseLong(next());
		 }

	}

}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

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

public class Main{
    //List l1=new ArrayList<>();
    public static void main(String args[]){
        Scanner obj=new Scanner(System.in);
        int t=obj.nextInt();
        while(t-->0)soln(obj);
    }
    static void soln(Scanner obj){
        long mod=1000000007,cnt=0,flg=0;
        int x=0;
        String y="YES", o="NO";
        int n=obj.nextInt();
        int[] a=new int[n];
        for(int i=0;i<n;i++){
            a[i]=obj.nextInt();
        }
        Arrays.sort(a);
        x=a[n-1];
        for(int i=n-1;i>=0;i--){
            if(a[i]>=x)cnt++;
        }

        System.out.println(((cnt&1)==0)?o:y);
    }

    static Map<Long,Long> mapInit(Map<Long,Long> map,long x){
            if(map.containsKey(x))map.put(x,map.get(x)+1);
            else map.put(x,1L);
            return map;
    }
    public static boolean isSorted(int[] array) {
        return IntStream.range(0, array.length - 1).noneMatch(i -> array[i] > array[i + 1]);
    }
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on the logic of the game and how the code handles the array. The code sorts the array and then counts how many elements are greater than or equal to the maximum element in the array. However, this approach is flawed because the game involves a sequence of operations where the maximum value is updated and elements are set to zero.

A key insight is that the game is more complex than just counting the number of elements greater than or equal to the maximum. The correct approach involves understanding the game dynamics and the sequence of moves.

### Test Input to Expose the Bug

Consider the following test case:
- An array where the maximum value is repeated, and the sequence of moves can lead to a situation where the player who starts (Alice) can force a win or a loss based on the number of maximum values and their positions.

Here is a small-scale test input that should expose the bug:

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

### Explanation

1. **Test Case 1: `3 1 3 3`**
   - Alice can choose the second or third element (both are 3).
   - After Alice's move, the array becomes `[1, 0, 3]` or `[1, 3, 0]` with `mx = 3`.
   - Bob can then choose the remaining 3.
   - After Bob's move, the array becomes `[1, 0, 0]` with `mx = 3`.
   - Alice can choose the 1.
   - After Alice's move, the array becomes `[0, 0, 0]` with `mx = 3`.
   - Bob can no longer make a move.
   - **Alice wins.**

2. **Test Case 2: `3 3 3 1`**
   - Alice can choose any of the three 3s.
   - After Alice's move, the array becomes `[3, 3, 0]` with `mx = 3`.
   - Bob can choose any of the remaining 3s.
   - After Bob's move, the array becomes `[3, 0, 0]` with `mx = 3`.
   - Alice can choose the remaining 3.
   - After Alice's move, the array becomes `[0, 0, 0]` with `mx = 3`.
   - Bob can no longer make a move.
   - **Alice wins.**

3. **Test Case 3: `4 2 2 2 3`**
   - Alice can choose the fourth element (3).
   - After Alice's move, the array becomes `[2, 2, 2, 0]` with `mx = 3`.
   - Bob can choose any of the three 2s.
   - After Bob's move, the array becomes `[2, 2, 0, 0]` with `mx = 3`.
   - Alice can choose any of the remaining 2s.
   - After Alice's move, the array becomes `[2, 0, 0, 0]` with `mx = 3`.
   - Bob can choose the remaining 2.
   - After Bob's move, the array becomes `[0, 0, 0, 0]` with `mx = 3`.
   - Alice can no longer make a move.
   - **Bob wins.**

This test input should expose the bug in the code, as the current logic in the code does not correctly handle the game dynamics and may incorrectly determine the winner.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <queue>

using namespace std;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vll = vector<ll>;
ll n, m, k, l, M = 1000000007;

template <typename T>
static istream& operator>>(std::istream& in, std::vector<T>& vec) {
    for (auto& elem : vec) {
        in >> elem;
    }
    return in;
}
static void sort(vector<int>& vec) {
    sort(vec.begin(), vec.end());
}

class karim
{
public:
    void solve()
    {
        bool flag = true;
        cin >> n;
		vector<int> a(n);
		cin >> a;
        sort(a.begin(), a.end(), greater<int>());
        int mx = 0;
        for (int i = 0; i < n; i++)
        {
			if (a[i] >= mx)
            {
				mx = a[i];
				a[i] = 0;
				flag = !flag;
			}
        }
		if (!flag)
		{
			cout << "YES" << endl;
		}
		else
		{
			cout << "NO" << endl;
		}
    }
};
int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    karim solution;
    int t = 1;
    cin >> t;
    while (t--) {
        solution.solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on edge cases where the sorting and the logic for determining the winner might fail. The key points to consider are:

1. **Arrays with all elements equal**: This can expose issues with the logic if the algorithm incorrectly assumes that the first player can always win.
2. **Arrays where the first few elements are the same and then differ**: This can test the sorting and the turn-taking logic.
3. **Arrays with a mix of high and low values**: This can test the logic for updating `mx` and determining the winner.

Let's consider a test case where the array has a mix of high and low values, and the first few elements are the same:

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

### Explanation:
- **Test Case**: 1
- **Array Size**: 4
- **Array Elements**: [2, 2, 3, 1]

### Reasoning:
- **Initial State**: `a = [2, 2, 3, 1]`, `mx = 0`
- **Alice's Turn**:
  - Alice can choose any of the indices 1, 2, or 3 (0-based indexing) since `a[1] = 2`, `a[2] = 2`, and `a[3] = 3` are all greater than or equal to `mx = 0`.
  - Suppose Alice chooses `i = 3` (0-based indexing). Then `a = [2, 2, 0, 1]` and `mx = 3`.
- **Bob's Turn**:
  - Bob can choose `i = 1` or `i = 2` (0-based indexing) since `a[1] = 2` and `a[2] = 2` are both greater than or equal to `mx = 3`.
  - Suppose Bob chooses `i = 1` (0-based indexing). Then `a = [2, 0, 0, 1]` and `mx = 2`.
- **Alice's Turn**:
  - Alice can choose `i = 0` (0-based indexing) since `a[0] = 2` is greater than or equal to `mx = 2`.
  - Suppose Alice chooses `i = 0` (0-based indexing). Then `a = [0, 0, 0, 1]` and `mx = 2`.
- **Bob's Turn**:
  - Bob can choose `i = 3` (0-based indexing) since `a[3] = 1` is greater than or equal to `mx = 2`.
  - Suppose Bob chooses `i = 3` (0-based indexing). Then `a = [0, 0, 0, 0]` and `mx = 1`.
- **Alice's Turn**:
  - Alice cannot make any move since all elements are 0.

In this scenario, Alice does not have a winning strategy. The expected output is "NO". This test case should expose the bug in the buggy code, which might incorrectly assume that Alice can always win in such scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
int t,n,a[55],f[55];
int main(){
	cin>>t;
	while(t--){
		memset(f,0,sizeof(f));
		int maxn=0;
		cin>>n;
		for(int i=1;i<=n;++i){
			cin>>a[i];
			maxn=max(maxn,a[i]);
			f[a[i]]++;
		}
		if(f[maxn]%2==1||n%2==1){
			cout<<"YES\n";
		}
		else{
			cout<<"NO\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
5
2
2 2
3
3 2 1
4
4 3 2 1
5
5 4 3 2 1
6
6 5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
N = int(input())

for i in range(N):
    length = int(input())
    arr = [int(i) for i in input().split()]
    arr.sort()
    num = arr.count(arr[-1])
    if num % 2 == 0:
        print('NO')
    else:
        print('YES')

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
t=int(input())
for i in range(0,t):
    n=int(input())
    a=input()
    arr=list(map(int,a.split()))
    arr.sort()
    max=arr[-1]
    count=0
    if max==0:
        if n%2!=0:
            print("yes")
        else:
            print("no")
    else:
        if n>2:
            for j in range(n-2,-1,-1):
                if arr[j]==max:
                    count=count+1

            if count%2==0:
                print("yes")
            else:
                print("no")
        if n==2:
            if arr[0]!=arr[1]:
                print("yes")
            else:
                print("no")
        if n==1:
            print("yes")

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
//Modulus properties
 //1 (a+b)%y = (a%y+b%y)%y also
// <if to find pairs st (a+b)%y==0 one is (a%y) and other is b%y==((y-(a%y))%y)>
 // < if (a+b)%y == (c+b)%y  then a%y = c%y (in  terms of remainder) >
// 2 (a-b)%y = (a%y-b%y+y)%y also
//   < if (a-b)%y ==0 then a%y = b%y  >
import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;
import java.util.LinkedList;
import java.util.Arrays;
import java.util.*;

public class one {

    static final int INF = Integer.MAX_VALUE;
    static final long MOD = 1000000007;

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

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

    static class Graph {
        int V;
        LinkedList<Integer>[] adjList;

        @SuppressWarnings("unchecked")
        Graph(int V) {
            this.V = V;
            adjList = new LinkedList[V];
            for (int i = 0; i < V; i++)
                adjList[i] = new LinkedList<>();
        }

        void addEdge(int u, int v) {
            adjList[u].add(v);
        }

        // Add more graph algorithms as needed
    }

    static class DSU {
        int[] parent;
        int[] rank;

        DSU(int n) {
            parent = new int[n];
            rank = new int[n];
            for (int i = 0; i < n; i++) {
                parent[i] = i;
                rank[i] = 0;
            }
        }

        int find(int x) {
            if (parent[x] != x)
                parent[x] = find(parent[x]);
            return parent[x];
        }

        void union(int x, int y) {
            int xRoot = find(x);
            int yRoot = find(y);
            if (xRoot == yRoot)
                return;
            if (rank[xRoot] < rank[yRoot])
                parent[xRoot] = yRoot;
            else if (rank[xRoot] > rank[yRoot])
                parent[yRoot] = xRoot;
            else {
                parent[yRoot] = xRoot;
                rank[xRoot]++;
            }
        }
    }

    static int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    static long gcd(long a, long b) {
        while (b != 0) {
            long temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }

    static long power(long x, long y, long p) {
        long res = 1;
        x = x % p;

        while (y > 0) {
            if (y % 2 == 1)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

    static boolean[] sieveOfEratosthenes(int n) {
        boolean[] prime = new boolean[n + 1];
        Arrays.fill(prime, true);
        prime[0] = prime[1] = false;
        for (int p = 2; p * p <= n; p++) {
            if (prime[p]) {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;
    }

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        int t = sc.nextInt();
        while (t > 0) {
            int n = sc.nextInt();
            int arr[] = new int[n];
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextInt();
            }
           Arrays.sort(arr);
           int c = 0;
           int p = arr[n-1];
           for (int i = 0; i < n; i++) {
            if(arr[i]==p)c++;
           }
           if(c%2==0){
            System.out.println("NO");
           }
              else{
                System.out.println("YES");
              }
   t--;
        }
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given code, we need to focus on edge cases or scenarios where the code's logic might fail. The code sorts the array and checks if the maximum value appears an odd number of times to determine if Alice has a winning strategy. However, this approach is flawed because it doesn't consider the game's dynamics correctly.

### Step-by-Step Analysis:
1. **Understanding the Bug:**
   - The code sorts the array and checks if the maximum value appears an odd number of times.
   - This logic is incorrect because the game involves players taking turns and the state of the array changes after each operation.

2. **Identifying Edge Cases:**
   - **Single Maximum Value:** If there is only one maximum value, Alice can always win by choosing it first.
   - **Multiple Maximum Values:** If there are multiple maximum values, the game dynamics can be more complex.
   - **Equal Elements:** If all elements are equal, the game can be a draw or a win depending on the number of elements.

3. **Creating a Test Case:**
   - We need a test case where the code's logic fails to determine the correct winner.
   - A good candidate is an array where the maximum value appears an even number of times, but Alice still has a winning strategy due to the game dynamics.

### Chosen Test Case:
- **Array with Multiple Maximum Values:** Consider an array where the maximum value appears an even number of times, but Alice can still win by strategically choosing elements.

### Example:
```plaintext
1
4
2 2 3 3
```

### Explanation:
- **Array:** [2, 2, 3, 3]
- **Maximum Value:** 3
- **Occurrences of Maximum Value:** 2 (even)

**Game Dynamics:**
1. Alice can choose one of the 3s.
2. After Alice's move, the array becomes [2, 2, 0, 3] and `mx` is 3.
3. Bob can choose the remaining 3.
4. After Bob's move, the array becomes [2, 2, 0, 0] and `mx` is 3.
5. Alice can now choose one of the 2s.
6. After Alice's move, the array becomes [0, 2, 0, 0] and `mx` is 3.
7. Bob can choose the remaining 2.
8. After Bob's move, the array becomes [0, 0, 0, 0] and `mx` is 3.
9. Alice can no longer make a move, but Bob can't either, so the game ends in a draw.

However, in the context of the problem, if the game ends in a draw, it is considered a loss for Alice. Therefore, the correct output should be "NO".

### Expected Output:
```plaintext
NO
```

### Actual Output of the Buggy Code:
- The buggy code will output "NO" because it sees that the maximum value 3 appears an even number of times.

### Conclusion:
This test case does not expose the bug because the correct output matches the buggy code's output. We need a different test case where the correct output is "YES" but the buggy code outputs "NO".

### Correct Test Case:
```plaintext
1
4
2 3 3 4
```

### Explanation:
- **Array:** [2, 3, 3, 4]
- **Maximum Value:** 4
- **Occurrences of Maximum Value:** 1 (odd)

**Game Dynamics:**
1. Alice can choose the 4.
2. After Alice's move, the array becomes [2, 3, 3, 0] and `mx` is 4.
3. Bob can choose one of the 3s.
4. After Bob's move, the array becomes [2, 0, 3, 0] and `mx` is 4.
5. Alice can choose the remaining 3.
6. After Alice's move, the array becomes [2, 0, 0, 0] and `mx` is 4.
7. Bob can choose the 2.
8. After Bob's move, the array becomes [0, 0, 0, 0] and `mx` is 4.
9. Alice can no longer make a move, but Bob can't either, so the game ends in a draw.

However, in the context of the problem, if the game ends in a draw, it is considered a loss for Alice. Therefore, the correct output should be "YES".

### Expected Output:
```plaintext
YES
```

### Actual Output of the Buggy Code:
- The buggy code will output "YES" because it sees that the maximum value 4 appears an odd number of times.

### Final Test Case:
```plaintext
1
4
2 3 3 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Submission Bait
time_limit_ms: 1000
memory_limit_mb: 256
Description: Alice and Bob are playing a game in an array $a$ of size $n$.

They take turns to do operations, with Alice starting first. The player who can not operate will lose. At first, a variable $mx$ is set to $0$.

In one operation, a player can do:

  * Choose an index $i$ ($1 \le i \le n$) such that $a_{i} \geq mx$ and set $mx$ to $a_{i}$. Then, set $a_{i}$ to $0$.

Determine whether Alice has a winning strategy.

Input: The first line contains an integer $t$ ($1 \leq t \leq 10^3$) — the number of test cases.

For each test case:

  * The first line contains an integer $n$ ($2 \leq n \leq 50$) — the size of the array.    * The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq n$) — the elements of the array.

Output: For each test case, if Alice has a winning strategy, output "YES". Otherwise, output "NO".

You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.

Examples:
input:
5
2
2 1
2
1 1
3
3 3 3
4
3 3 4 4
4
1 2 2 2
output:
YES
NO
YES
NO
YES

Note: In the first test case, Alice can choose $i=1$ since $a_1=2 \ge mx=0$.

After Alice's operation, $a=[0,1]$ and $mx=2$. Bob can not do any operation. Alice wins.

In the second test case, Alice doesn't have a winning strategy.

For example, if Alice chooses $i=1$, after Alice's operation: $a=[0,1]$ and $mx=1$. Then, Bob can choose $i=2$ since $a_2=1 \ge mx=1$. After Bob's operation: $a=[0,0]$ and $mx=1$. Alice can not do any operation. Bob wins.

Buggy Code:
import java.util.*;
import java.io.*;
public class SubmissionBait {
	public static void main(String [] args) {
		try {
			System.setIn(new FileInputStream("input.txt"));
			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
		} catch (Exception e) {
			System.err.println("Error");
		}

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

			HashMap<Integer , Integer> hm = new HashMap<>();
			for (int i : a) {
				int nq = hm.getOrDefault(i , 0);
				nq++;
				hm.put(i , nq);
			}
			// System.out.println(hm);
			// System.out.println(max);
			if (hm.get(max) % 2 == 1) {
				System.out.println("YES");
			} else {
				if (n % 2 == 1) {
					System.out.println("YES");
				} else {
					System.out.println("NO");
				}
			}

		}

		sc.close();
	}
	static int[] arrayGen(int size , int limit) {
		int [] ans = new int[size];
		for (int i = 0 ; i < size ; i++) {
			ans[i] = (int)(Math.random() * limit);
		}
		return ans;
	}

	static List<Integer> seive(int N) {
		boolean[] sieve = new boolean[N];

		for (int i = 0; i < sieve.length; i++) {
			sieve[i] = true;
		}
		for (int i = 2; i < Math.sqrt(N); i++) {
			if (sieve[i] == true) {
				for (int j = (i * i); j < N; j = j + i) {
					sieve[j] = false;
				}
			}
		}
		List<Integer> ls = new ArrayList<>();
		for (int i = 2; i < sieve.length; i++) {
			if (sieve[i] == true) {
				ls.add(i);
			}
		}
		return ls;
	}

	public static void swapInArray(int ind1 , int ind2 , int[] arr) {
		int temp = arr[ind1];
		arr[ind1] = arr[ind2];
		arr[ind2] = temp;
	}

	public static void printArray(int [] a) {
		for (int i : a) {
			System.out.print(i + " ");
		}
		System.out.println();
	}

	public static void printList(List<Integer> ls) {
		for (Integer i : ls) {
			System.out.print(i + " ");
		}
		System.out.println();
	}
	public static int gcd(int a , int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b , a % b);
	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on scenarios where the logic of the algorithm might fail. The buggy code has a specific condition that checks if the maximum value in the array appears an odd number of times. If it does, Alice is assumed to have a winning strategy. However, this logic is not always correct, especially in cases where the maximum value appears an even number of times but the array structure allows Alice to still have a winning strategy.

Let's consider the following edge cases:
1. **All elements are the same and even in number**: This is a straightforward case where the buggy code will incorrectly output "NO" because it checks the number of maximum elements.
2. **Multiple maximum values with an even count, but a strategic position allows Alice to win**: This is a more complex case where the structure of the array allows Alice to force Bob into a losing position.

### Test Input:
```plaintext
3
2
2 2
4
2 2 2 2
5
3 3 3 3 3
```

### Explanation:
1. **Test Case 1:**
   - Array: [2, 2]
   - The maximum value 2 appears twice (even), but Alice can still win by setting `mx` to 2 and making the array [0, 2]. Bob then sets `mx` to 2 and makes the array [0, 0]. Alice wins.

2. **Test Case 2:**
   - Array: [2, 2, 2, 2]
   - The maximum value 2 appears four times (even), but Alice can still win by setting `mx` to 2 and making the array [0, 2, 2, 2]. Bob then sets `mx` to 2 and makes the array [0, 0, 2, 2]. Alice then sets `mx` to 2 and makes the array [0, 0, 0, 2]. Bob then sets `mx` to 2 and makes the array [0, 0, 0, 0]. Alice wins.

3. **Test Case 3:**
   - Array: [3, 3, 3, 3, 3]
   - The maximum value 3 appears five times (odd), so the buggy code will output "YES". This is correct because Alice can always force Bob into a losing position by setting `mx` to 3 and making the array [0, 3, 3, 3, 3]. Bob then sets `mx` to 3 and makes the array [0, 0, 3, 3, 3]. Alice then sets `mx` to 3 and makes the array [0, 0, 0, 3, 3]. Bob then sets `mx` to 3 and makes the array [0, 0, 0, 0, 3]. Alice then sets `mx` to 3 and makes the array [0, 0, 0, 0, 0]. Alice wins.

This test input should expose the bug in the given code by providing scenarios where the maximum value appears an even number of times but Alice still has a winning strategy.