#include <bits/stdc++.h>
using namespace std;
mt19937 ran(time(0));
long long mod = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long Bpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans *= a;
      ans %= mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
map<pair<long long, long long>, long long> pos;
void P(long long i, long long j, long long k,
       vector<pair<long long, long long>>& a) {
  pair<long long, long long> t = a[k];
  a[k] = a[j];
  a[j] = a[i];
  a[i] = t;
  pos[a[i]] = i;
  pos[a[j]] = j;
  pos[a[k]] = k;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<pair<long long, long long>> a;
    map<long long, long long> now;
    bool ff = false;
    for (long long i = 0; i < n; i++) {
      long long b;
      cin >> b;
      a.push_back({b, now[b]++});
      if (now[b] == 2) {
        ff = true;
      }
    }
    if (ff) {
      cout << "YES\n";
      continue;
    }
    vector<pair<long long, long long>> b = a;
    sort(b.begin(), b.end());
    for (long long i = 0; i < n; i++) {
      pos[a[i]] = i;
    }
    bool f = false;
    for (long long i = 0; i < n; i++) {
      if (a[i] == b[i]) continue;
      if (pos[b[i]] != i + 1) {
        P(i, i + 1, pos[b[i]], a);
      } else {
        if (i + 1 == n - 1) {
          if (a[i].first == a[i + 1].first) break;
          if (a[i + 1].second) {
            P(i, i + 1, pos[{a[i + 1].first, a[i + 1].second - 1}], a);
            break;
          }
          f = true;
          break;
        }
        P(i, i + 2, pos[b[i]], a);
      }
    }
    if (f)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
}

### Prompt

Create a solution in CPP for 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;
mt19937 ran(time(0));
long long mod = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long Bpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans *= a;
      ans %= mod;
    }
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
map<pair<long long, long long>, long long> pos;
void P(long long i, long long j, long long k,
       vector<pair<long long, long long>>& a) {
  pair<long long, long long> t = a[k];
  a[k] = a[j];
  a[j] = a[i];
  a[i] = t;
  pos[a[i]] = i;
  pos[a[j]] = j;
  pos[a[k]] = k;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<pair<long long, long long>> a;
    map<long long, long long> now;
    bool ff = false;
    for (long long i = 0; i < n; i++) {
      long long b;
      cin >> b;
      a.push_back({b, now[b]++});
      if (now[b] == 2) {
        ff = true;
      }
    }
    if (ff) {
      cout << "YES\n";
      continue;
    }
    vector<pair<long long, long long>> b = a;
    sort(b.begin(), b.end());
    for (long long i = 0; i < n; i++) {
      pos[a[i]] = i;
    }
    bool f = false;
    for (long long i = 0; i < n; i++) {
      if (a[i] == b[i]) continue;
      if (pos[b[i]] != i + 1) {
        P(i, i + 1, pos[b[i]], a);
      } else {
        if (i + 1 == n - 1) {
          if (a[i].first == a[i + 1].first) break;
          if (a[i + 1].second) {
            P(i, i + 1, pos[{a[i + 1].first, a[i + 1].second - 1}], a);
            break;
          }
          f = true;
          break;
        }
        P(i, i + 2, pos[b[i]], a);
      }
    }
    if (f)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
}

```