#include <bits/stdc++.h>
using namespace std;
const int N = 500010;
int a[N];
int n;
int par[N], sz[N];
vector<int> pos[N];
int pa(int x) {
  if (par[x] == x) return x;
  return par[x] = pa(par[x]);
}
void join(int x, int y) {
  int tmp1 = pa(x);
  int tmp2 = pa(y);
  if (tmp1 == tmp2) return;
  par[tmp1] = tmp2;
  sz[tmp2] += sz[tmp1];
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      par[i] = i;
      sz[i] = 1;
      pos[i].clear();
    }
    bool l = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      pos[a[i]].push_back(i);
      if (pos[a[i]].size() > 1) {
        l = 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (auto it : pos[i]) join(i, it);
    }
    int cnt = 0;
    for (int i = 1; i <= n; ++i) {
      if (pa(i) == i) {
        if (sz[i] % 2 == 0) {
          ++cnt;
        }
      }
    }
    if (cnt % 2 == 0 || l) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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;
const int N = 500010;
int a[N];
int n;
int par[N], sz[N];
vector<int> pos[N];
int pa(int x) {
  if (par[x] == x) return x;
  return par[x] = pa(par[x]);
}
void join(int x, int y) {
  int tmp1 = pa(x);
  int tmp2 = pa(y);
  if (tmp1 == tmp2) return;
  par[tmp1] = tmp2;
  sz[tmp2] += sz[tmp1];
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      par[i] = i;
      sz[i] = 1;
      pos[i].clear();
    }
    bool l = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", a + i);
      pos[a[i]].push_back(i);
      if (pos[a[i]].size() > 1) {
        l = 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (auto it : pos[i]) join(i, it);
    }
    int cnt = 0;
    for (int i = 1; i <= n; ++i) {
      if (pa(i) == i) {
        if (sz[i] % 2 == 0) {
          ++cnt;
        }
      }
    }
    if (cnt % 2 == 0 || l) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}

```