#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5e5 + 7;
const int maxn = 5e4 + 7;
const int inf = 1e6 + 10;
const double pp = 1e-6;
template <class type>
inline void read(type &k) {
  k = 0;
  type t = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    k = k * 10 + ch - '0';
    ch = getchar();
  }
  k *= t;
}
inline long long gcd(long long x, long long y) {
  return y == 0 ? x : gcd(y, x % y);
}
long long qpow(long long a, long long b) {
  long long aa = 1;
  while (b) {
    if (b % 2 == 1) aa = (aa * a);
    a = (a * a);
    b /= 2;
  }
  return aa;
}
int a[MAXN], b[MAXN], tmp[MAXN];
long long merge_sort(int q[], int l, int r) {
  if (l >= r) return 0;
  int mid = l + r >> 1;
  long long res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
  int k = 0, i = l, j = mid + 1;
  while (i <= mid && j <= r)
    if (q[i] <= q[j])
      tmp[k++] = q[i++];
    else {
      res += mid - i + 1;
      tmp[k++] = q[j++];
    }
  while (i <= mid) tmp[k++] = q[i++];
  while (j <= r) tmp[k++] = q[j++];
  for (i = l, j = 0; i <= r; i++, j++) q[i] = tmp[j];
  return res;
}
int main() {
  int t;
  read(t);
  while (t--) {
    int n;
    bool flag = false;
    read(n);
    for (int i = 1; i <= n; i++) {
      read(a[i]);
      b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    int now = 1;
    for (int i = 1; i <= n; i++) {
      if (i != 1 && b[i] == b[i - 1]) flag = true;
    }
    int ans = merge_sort(a, 1, n);
    printf((ans % 2) && !flag ? "NO\n" : "YES\n");
  }
}

### Prompt

Please formulate a CPP solution to 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 MAXN = 5e5 + 7;
const int maxn = 5e4 + 7;
const int inf = 1e6 + 10;
const double pp = 1e-6;
template <class type>
inline void read(type &k) {
  k = 0;
  type t = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') t = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    k = k * 10 + ch - '0';
    ch = getchar();
  }
  k *= t;
}
inline long long gcd(long long x, long long y) {
  return y == 0 ? x : gcd(y, x % y);
}
long long qpow(long long a, long long b) {
  long long aa = 1;
  while (b) {
    if (b % 2 == 1) aa = (aa * a);
    a = (a * a);
    b /= 2;
  }
  return aa;
}
int a[MAXN], b[MAXN], tmp[MAXN];
long long merge_sort(int q[], int l, int r) {
  if (l >= r) return 0;
  int mid = l + r >> 1;
  long long res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
  int k = 0, i = l, j = mid + 1;
  while (i <= mid && j <= r)
    if (q[i] <= q[j])
      tmp[k++] = q[i++];
    else {
      res += mid - i + 1;
      tmp[k++] = q[j++];
    }
  while (i <= mid) tmp[k++] = q[i++];
  while (j <= r) tmp[k++] = q[j++];
  for (i = l, j = 0; i <= r; i++, j++) q[i] = tmp[j];
  return res;
}
int main() {
  int t;
  read(t);
  while (t--) {
    int n;
    bool flag = false;
    read(n);
    for (int i = 1; i <= n; i++) {
      read(a[i]);
      b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    int now = 1;
    for (int i = 1; i <= n; i++) {
      if (i != 1 && b[i] == b[i - 1]) flag = true;
    }
    int ans = merge_sort(a, 1, n);
    printf((ans % 2) && !flag ? "NO\n" : "YES\n");
  }
}

```