#include <bits/stdc++.h>
using namespace std;
const long long sz = 5e5 + 10;
long long y[sz];
inline long long read() {
  long long f = 1, x = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  return f * x;
}
long long ans, n;
long long c[sz], a[sz];
void msort(long long b, long long e) {
  if (b == e) return;
  long long mid = (b + e) / 2, i = b, j = mid + 1, k = b;
  msort(b, mid), msort(mid + 1, e);
  while (i <= mid && j <= e)
    if (a[i] <= a[j])
      c[k++] = a[i++];
    else
      c[k++] = a[j++], ans += mid - i + 1;
  while (i <= mid) c[k++] = a[i++];
  while (j <= e) c[k++] = a[j++];
  for (long long l = b; l <= e; l++) a[l] = c[l];
}
bool z[sz];
void solve() {
  long long n = read();
  memset(z, 0, sizeof(bool) * (n + 2));
  bool f = 0;
  for (long long i = 1; i <= n; i++) {
    a[i] = read();
    if (!z[a[i]])
      z[a[i]] = 1;
    else
      f = 1;
  }
  ans = 0;
  msort(1, n);
  if (ans & 1) {
    if (f)
      puts("YES");
    else
      puts("NO");
  } else
    puts("YES");
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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 long long sz = 5e5 + 10;
long long y[sz];
inline long long read() {
  long long f = 1, x = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  return f * x;
}
long long ans, n;
long long c[sz], a[sz];
void msort(long long b, long long e) {
  if (b == e) return;
  long long mid = (b + e) / 2, i = b, j = mid + 1, k = b;
  msort(b, mid), msort(mid + 1, e);
  while (i <= mid && j <= e)
    if (a[i] <= a[j])
      c[k++] = a[i++];
    else
      c[k++] = a[j++], ans += mid - i + 1;
  while (i <= mid) c[k++] = a[i++];
  while (j <= e) c[k++] = a[j++];
  for (long long l = b; l <= e; l++) a[l] = c[l];
}
bool z[sz];
void solve() {
  long long n = read();
  memset(z, 0, sizeof(bool) * (n + 2));
  bool f = 0;
  for (long long i = 1; i <= n; i++) {
    a[i] = read();
    if (!z[a[i]])
      z[a[i]] = 1;
    else
      f = 1;
  }
  ans = 0;
  msort(1, n);
  if (ans & 1) {
    if (f)
      puts("YES");
    else
      puts("NO");
  } else
    puts("YES");
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```