#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const long long INF = 1e18;
const long long N = 500005;
int t[4 * N];
void update(int v, int l, int r, int pos) {
  if (l == r) {
    t[v] = 1 - t[v];
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m) {
    update(2 * v, l, m, pos);
  } else {
    update(2 * v + 1, m + 1, r, pos);
  }
  t[v] = t[2 * v] + t[2 * v + 1];
}
int get(int v, int l, int r, int s, int e) {
  if (s > e) {
    return 0;
  }
  if (l == s and r == e) {
    return t[v];
  }
  int m = (l + r) / 2;
  return get(2 * v, l, m, s, min(m, e)) +
         get(2 * v + 1, m + 1, r, max(m + 1, s), e);
}
bool b[N];
long long x[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    bool check = false;
    long long ans = 0;
    for (long long i = (1); i < (n + 1); i++) {
      b[i] = 0;
    }
    for (long long i = (1); i < (n + 1); i++) {
      cin >> x[i];
      if (b[x[i]]) {
        check = true;
      }
      b[x[i]] = 1;
    }
    if (check) {
      cout << "YES" << endl;
      continue;
    }
    update(1, 1, n, x[1]);
    for (long long i = (2); i < (n + 1); i++) {
      ans += get(1, 1, n, x[i] + 1, n);
      update(1, 1, n, x[i]);
    }
    for (long long i = (1); i < (n + 1); i++) {
      update(1, 1, n, x[i]);
    }
    cout << "NO\0YES" + 3 * (ans % 2 == 0) << endl;
  }
  return 0;
}

### Prompt

Generate 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 long long MOD = 998244353;
const long long INF = 1e18;
const long long N = 500005;
int t[4 * N];
void update(int v, int l, int r, int pos) {
  if (l == r) {
    t[v] = 1 - t[v];
    return;
  }
  int m = (l + r) / 2;
  if (pos <= m) {
    update(2 * v, l, m, pos);
  } else {
    update(2 * v + 1, m + 1, r, pos);
  }
  t[v] = t[2 * v] + t[2 * v + 1];
}
int get(int v, int l, int r, int s, int e) {
  if (s > e) {
    return 0;
  }
  if (l == s and r == e) {
    return t[v];
  }
  int m = (l + r) / 2;
  return get(2 * v, l, m, s, min(m, e)) +
         get(2 * v + 1, m + 1, r, max(m + 1, s), e);
}
bool b[N];
long long x[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int tt;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    bool check = false;
    long long ans = 0;
    for (long long i = (1); i < (n + 1); i++) {
      b[i] = 0;
    }
    for (long long i = (1); i < (n + 1); i++) {
      cin >> x[i];
      if (b[x[i]]) {
        check = true;
      }
      b[x[i]] = 1;
    }
    if (check) {
      cout << "YES" << endl;
      continue;
    }
    update(1, 1, n, x[1]);
    for (long long i = (2); i < (n + 1); i++) {
      ans += get(1, 1, n, x[i] + 1, n);
      update(1, 1, n, x[i]);
    }
    for (long long i = (1); i < (n + 1); i++) {
      update(1, 1, n, x[i]);
    }
    cout << "NO\0YES" + 3 * (ans % 2 == 0) << endl;
  }
  return 0;
}

```