#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 5e5 + 5;
const int M = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3fll;
const double pi = acos(-1.0);
int read() {
  int a = 0;
  int d = 1;
  char c;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') d = -1;
  a = a * 10 + c - 48;
  while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
  a *= d;
  return a;
}
int a[N];
int c[N];
map<int, bool> mp;
int ans;
;
void msort(int l, int r) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  msort(l, mid);
  msort(mid + 1, r);
  int i = l;
  int j = mid + 1;
  int idx = l;
  while (i <= mid && j <= r) {
    if (a[i] <= a[j])
      c[idx++] = a[i++];
    else {
      ans += mid - i + 1;
      c[idx++] = a[j++];
    }
  }
  while (i <= mid) c[idx++] = a[i++];
  while (j <= r) c[idx++] = a[j++];
  for (int k = l; k <= r; k++) a[k] = c[k];
}
signed main() {
  int t = read();
  while (t--) {
    mp.clear();
    int n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    bool f1 = false;
    bool f2 = false;
    for (int i = 1; i <= n; i++) {
      if (a[i] < a[i - 1]) f1 = true;
      if (mp[a[i]]) {
        f2 = true;
        break;
      }
      mp[a[i]] = true;
    }
    if (!f1 || f2) {
      puts("YES");
      continue;
    }
    if (n < 3) {
      puts("NO");
      continue;
    }
    ans = 0;
    msort(1, n);
    if (ans % 2 == 0)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 5e5 + 5;
const int M = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3fll;
const double pi = acos(-1.0);
int read() {
  int a = 0;
  int d = 1;
  char c;
  while (c = getchar(), c < '0' || c > '9')
    if (c == '-') d = -1;
  a = a * 10 + c - 48;
  while (c = getchar(), c >= '0' && c <= '9') a = a * 10 + c - 48;
  a *= d;
  return a;
}
int a[N];
int c[N];
map<int, bool> mp;
int ans;
;
void msort(int l, int r) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  msort(l, mid);
  msort(mid + 1, r);
  int i = l;
  int j = mid + 1;
  int idx = l;
  while (i <= mid && j <= r) {
    if (a[i] <= a[j])
      c[idx++] = a[i++];
    else {
      ans += mid - i + 1;
      c[idx++] = a[j++];
    }
  }
  while (i <= mid) c[idx++] = a[i++];
  while (j <= r) c[idx++] = a[j++];
  for (int k = l; k <= r; k++) a[k] = c[k];
}
signed main() {
  int t = read();
  while (t--) {
    mp.clear();
    int n = read();
    for (int i = 1; i <= n; i++) a[i] = read();
    bool f1 = false;
    bool f2 = false;
    for (int i = 1; i <= n; i++) {
      if (a[i] < a[i - 1]) f1 = true;
      if (mp[a[i]]) {
        f2 = true;
        break;
      }
      mp[a[i]] = true;
    }
    if (!f1 || f2) {
      puts("YES");
      continue;
    }
    if (n < 3) {
      puts("NO");
      continue;
    }
    ans = 0;
    msort(1, n);
    if (ans % 2 == 0)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}

```