#include <bits/stdc++.h>
using namespace std;
void up(long long i, long long x, long long lx, long long rx,
        vector<long long> &tree) {
  if (rx - lx == 1) {
    tree[x]++;
  } else {
    long long helpp = (rx + lx) / 2;
    if (i < helpp) {
      up(i, 2 * x + 1, lx, helpp, tree);
    } else {
      up(i, 2 * x + 2, helpp, rx, tree);
    }
    tree[x] = tree[2 * x + 1] + tree[2 * x + 2];
  }
  return;
}
long long gan(long long x, long long l, long long r, long long lx, long long rx,
              vector<long long> &tree) {
  if (lx >= l and rx <= r) {
    return tree[x];
  } else if (lx >= r or l >= rx) {
    return 0;
  } else {
    long long helpp = (rx + lx) / 2;
    return gan(2 * x + 1, l, r, lx, helpp, tree) +
           gan(2 * x + 2, l, r, helpp, rx, tree);
  }
}
signed main() {
  long long amm;
  cin >> amm;
  while (amm--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    vector<long long> ch(n);
    long long anss = 0;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      a[i]--;
      ch[a[i]]++;
      if (ch[a[i]] > 1) {
        anss = 1;
      }
    }
    if (anss == 1) {
      cout << "YES\n";
    } else {
      long long sz = 1;
      while (sz <= n) {
        sz *= 2;
      }
      vector<long long> st(sz * 2 - 1);
      long long ans = 0;
      for (long long i = 0; i < n; i++) {
        ans += gan(0, a[i] + 1, n, 0, sz, st);
        up(a[i], 0, 0, sz, st);
      }
      if (ans % 2 == 0) {
        cout << "YES\n";
      } else {
        cout << "NO\n";
      }
    }
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

Petya has an array of integers a_1, a_2, …, a_n. He only likes sorted arrays. Unfortunately, the given array could be arbitrary, so Petya wants to sort it.

Petya likes to challenge himself, so he wants to sort array using only 3-cycles. More formally, in one operation he can pick 3 pairwise distinct indices i, j, and k (1 ≤ i, j, k ≤ n) and apply i → j → k → i cycle to the array a. It simultaneously places a_i on position j, a_j on position k, and a_k on position i, without changing any other element.

For example, if a is [10, 50, 20, 30, 40, 60] and he chooses i = 2, j = 1, k = 5, then the array becomes [\underline{50}, \underline{40}, 20, 30, \underline{10}, 60].

Petya can apply arbitrary number of 3-cycles (possibly, zero). You are to determine if Petya can sort his array a, i. e. make it non-decreasing.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 5 ⋅ 10^5). Description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the length of the array a.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n).

It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5.

Output

For each test case, print "YES" (without quotes) if Petya can sort the array a using 3-cycles, and "NO" (without quotes) otherwise. You can print each letter in any case (upper or lower).

Example

Input


7
1
1
2
2 2
2
2 1
3
1 2 3
3
2 1 3
3
3 1 2
4
2 1 4 3


Output


YES
YES
NO
YES
NO
YES
YES

Note

In the 6-th test case Petya can use the 3-cycle 1 → 3 → 2 → 1 to sort the array.

In the 7-th test case Petya can apply 1 → 3 → 2 → 1 and make a = [1, 4, 2, 3]. Then he can apply 2 → 4 → 3 → 2 and finally sort the array.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void up(long long i, long long x, long long lx, long long rx,
        vector<long long> &tree) {
  if (rx - lx == 1) {
    tree[x]++;
  } else {
    long long helpp = (rx + lx) / 2;
    if (i < helpp) {
      up(i, 2 * x + 1, lx, helpp, tree);
    } else {
      up(i, 2 * x + 2, helpp, rx, tree);
    }
    tree[x] = tree[2 * x + 1] + tree[2 * x + 2];
  }
  return;
}
long long gan(long long x, long long l, long long r, long long lx, long long rx,
              vector<long long> &tree) {
  if (lx >= l and rx <= r) {
    return tree[x];
  } else if (lx >= r or l >= rx) {
    return 0;
  } else {
    long long helpp = (rx + lx) / 2;
    return gan(2 * x + 1, l, r, lx, helpp, tree) +
           gan(2 * x + 2, l, r, helpp, rx, tree);
  }
}
signed main() {
  long long amm;
  cin >> amm;
  while (amm--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    vector<long long> ch(n);
    long long anss = 0;
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
      a[i]--;
      ch[a[i]]++;
      if (ch[a[i]] > 1) {
        anss = 1;
      }
    }
    if (anss == 1) {
      cout << "YES\n";
    } else {
      long long sz = 1;
      while (sz <= n) {
        sz *= 2;
      }
      vector<long long> st(sz * 2 - 1);
      long long ans = 0;
      for (long long i = 0; i < n; i++) {
        ans += gan(0, a[i] + 1, n, 0, sz, st);
        up(a[i], 0, 0, sz, st);
      }
      if (ans % 2 == 0) {
        cout << "YES\n";
      } else {
        cout << "NO\n";
      }
    }
  }
}

```