#include <bits/stdc++.h>
using namespace std;
const int DIM = 5e5 + 7;
const int INF = 7e18 + 7;
const int mod = 998244353;
const int base = 29;
int n, t, k;
int a[DIM];
int b[DIM];
int pos[DIM];
set<int> s[DIM];
void del(int type, int var) { s[type].erase(var); }
void clr(int type, int var) {
  while (s[type].size() && (*s[type].begin()) <= var)
    s[type].erase(s[type].begin());
}
void add(int type, int var) { s[type].insert(var); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    set<int> all;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      all.insert(a[i]);
      b[i] = a[i];
      s[a[i]].insert(i);
    }
    if (all.size() != n) {
      cout << "YES\n";
      for (int i = 1; i <= n; i++) {
        s[b[i]].clear();
      }
      continue;
    }
    sort(b + 1, b + n + 1);
    int good = 1;
    int el1, el2, el3, k1, k2, k3;
    for (int i = 1; i <= n; i++) {
      if (b[i] == a[i]) {
        clr(a[i], i);
        continue;
      }
      if (i == n - 1) {
        good = 0;
        break;
      };
      el1 = a[i];
      k1 = i;
      el2 = b[i];
      k2 = (*s[el2].begin());
      el3 = a[i + 1];
      if (k2 == i + 1) el3 = a[i + 2];
      k3 = (*s[el3].rbegin());
      a[k1] = el2;
      a[k2] = el3;
      a[k3] = el1;
      del(el1, k1);
      del(el2, k2);
      del(el3, k3);
      add(el2, k1);
      add(el3, k2);
      add(el1, k3);
      clr(a[i], i);
    }
    for (int i = 1; i <= n; i++) {
      s[b[i]].clear();
    }
    if (good)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 DIM = 5e5 + 7;
const int INF = 7e18 + 7;
const int mod = 998244353;
const int base = 29;
int n, t, k;
int a[DIM];
int b[DIM];
int pos[DIM];
set<int> s[DIM];
void del(int type, int var) { s[type].erase(var); }
void clr(int type, int var) {
  while (s[type].size() && (*s[type].begin()) <= var)
    s[type].erase(s[type].begin());
}
void add(int type, int var) { s[type].insert(var); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    set<int> all;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      all.insert(a[i]);
      b[i] = a[i];
      s[a[i]].insert(i);
    }
    if (all.size() != n) {
      cout << "YES\n";
      for (int i = 1; i <= n; i++) {
        s[b[i]].clear();
      }
      continue;
    }
    sort(b + 1, b + n + 1);
    int good = 1;
    int el1, el2, el3, k1, k2, k3;
    for (int i = 1; i <= n; i++) {
      if (b[i] == a[i]) {
        clr(a[i], i);
        continue;
      }
      if (i == n - 1) {
        good = 0;
        break;
      };
      el1 = a[i];
      k1 = i;
      el2 = b[i];
      k2 = (*s[el2].begin());
      el3 = a[i + 1];
      if (k2 == i + 1) el3 = a[i + 2];
      k3 = (*s[el3].rbegin());
      a[k1] = el2;
      a[k2] = el3;
      a[k3] = el1;
      del(el1, k1);
      del(el2, k2);
      del(el3, k3);
      add(el2, k1);
      add(el3, k2);
      add(el1, k3);
      clr(a[i], i);
    }
    for (int i = 1; i <= n; i++) {
      s[b[i]].clear();
    }
    if (good)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}

```