#include <bits/stdc++.h>
using namespace std;
const int NR = 5e5 + 10;
int T, n;
int a[NR], num[NR];
struct arr {
  int mx, pos;
} t[NR * 4];
void build(int x, int lx, int rx) {
  if (lx == rx) {
    t[x].mx = a[lx];
    t[x].pos = lx;
    return;
  }
  int mid = (lx + rx) >> 1;
  build(x * 2, lx, mid);
  build(x * 2 + 1, mid + 1, rx);
  if (t[x * 2].mx > t[x * 2 + 1].mx)
    t[x] = t[x * 2];
  else
    t[x] = t[x * 2 + 1];
}
void add(int x, int lx, int rx, int p, int k) {
  if (lx == rx) {
    t[x].mx = k;
    t[x].pos = lx;
    return;
  }
  int mid = (lx + rx) >> 1;
  if (p <= mid)
    add(x * 2, lx, mid, p, k);
  else
    add(x * 2 + 1, mid + 1, rx, p, k);
  if (t[x * 2].mx > t[x * 2 + 1].mx)
    t[x] = t[x * 2];
  else
    t[x] = t[x * 2 + 1];
}
int main() {
  scanf("%d", &T);
  while (T--) {
    int flag = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) num[i] = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      num[a[i]]++;
      if (num[a[i]] >= 2) flag = 1;
    }
    if (flag) {
      printf("YES\n");
      continue;
    }
    build(1, 1, n);
    flag = 1;
    for (int i = n; i >= 1; i--) {
      int x = t[1].pos, y = 1;
      if (x == i) {
        add(1, 1, n, i, 0);
        continue;
      }
      if (x == 1 && i == 1) break;
      if (x == 1 && i == 2) {
        flag = 0;
        break;
      }
      if (x == 1) y = 2;
      int tmp = a[x];
      add(1, 1, n, x, a[y]);
      add(1, 1, n, y, a[i]);
      add(1, 1, n, i, 0);
      a[x] = a[y];
      a[y] = a[i];
      a[i] = tmp;
    }
    if (flag)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const int NR = 5e5 + 10;
int T, n;
int a[NR], num[NR];
struct arr {
  int mx, pos;
} t[NR * 4];
void build(int x, int lx, int rx) {
  if (lx == rx) {
    t[x].mx = a[lx];
    t[x].pos = lx;
    return;
  }
  int mid = (lx + rx) >> 1;
  build(x * 2, lx, mid);
  build(x * 2 + 1, mid + 1, rx);
  if (t[x * 2].mx > t[x * 2 + 1].mx)
    t[x] = t[x * 2];
  else
    t[x] = t[x * 2 + 1];
}
void add(int x, int lx, int rx, int p, int k) {
  if (lx == rx) {
    t[x].mx = k;
    t[x].pos = lx;
    return;
  }
  int mid = (lx + rx) >> 1;
  if (p <= mid)
    add(x * 2, lx, mid, p, k);
  else
    add(x * 2 + 1, mid + 1, rx, p, k);
  if (t[x * 2].mx > t[x * 2 + 1].mx)
    t[x] = t[x * 2];
  else
    t[x] = t[x * 2 + 1];
}
int main() {
  scanf("%d", &T);
  while (T--) {
    int flag = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) num[i] = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      num[a[i]]++;
      if (num[a[i]] >= 2) flag = 1;
    }
    if (flag) {
      printf("YES\n");
      continue;
    }
    build(1, 1, n);
    flag = 1;
    for (int i = n; i >= 1; i--) {
      int x = t[1].pos, y = 1;
      if (x == i) {
        add(1, 1, n, i, 0);
        continue;
      }
      if (x == 1 && i == 1) break;
      if (x == 1 && i == 2) {
        flag = 0;
        break;
      }
      if (x == 1) y = 2;
      int tmp = a[x];
      add(1, 1, n, x, a[y]);
      add(1, 1, n, y, a[i]);
      add(1, 1, n, i, 0);
      a[x] = a[y];
      a[y] = a[i];
      a[i] = tmp;
    }
    if (flag)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

```