#include <bits/stdc++.h>
using namespace std;
const int maxn = 500000 + 50;
int T, n, h[maxn], f[maxn], tot, ti[maxn];
int ans, sum;
int main() {
  scanf("%d", &T);
  for (int oo = 1; oo <= T; ++oo) {
    scanf("%d", &n);
    tot = 0;
    vector<bool> used(n);
    vector<set<int>> gi(n), ed(n);
    vector<int> seet;
    ans = 0, sum = 0;
    for (int i = 0; i < n; i++) f[i] = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &h[i]);
      tot = max(tot, ++f[--h[i]]);
    }
    for (int i = 0; i < n; ++i) ti[i] = h[i];
    sort(h, h + n);
    for (int i = 0; i < n; i++)
      if (h[i] != ti[i])
        gi[ti[i]].insert(i), seet.push_back(i), ed[h[i]].insert(i);
    function<void(int)> dfs = [&](int u) {
      ed[h[u]].erase(u);
      sum++;
      gi[ti[u]].erase(u);
      if (!ed[ti[u]].empty()) dfs(*ed[ti[u]].begin());
    };
    for (int i : seet)
      if (gi[ti[i]].count(i)) {
        sum = 0;
        dfs(i);
        if (sum % 2 == 0) ++ans;
      }
    bool p = 1;
    if (tot > 1 || ans % 2 == 0) p = 0;
    if (!p)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

### Prompt

Your challenge is to write 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 int maxn = 500000 + 50;
int T, n, h[maxn], f[maxn], tot, ti[maxn];
int ans, sum;
int main() {
  scanf("%d", &T);
  for (int oo = 1; oo <= T; ++oo) {
    scanf("%d", &n);
    tot = 0;
    vector<bool> used(n);
    vector<set<int>> gi(n), ed(n);
    vector<int> seet;
    ans = 0, sum = 0;
    for (int i = 0; i < n; i++) f[i] = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &h[i]);
      tot = max(tot, ++f[--h[i]]);
    }
    for (int i = 0; i < n; ++i) ti[i] = h[i];
    sort(h, h + n);
    for (int i = 0; i < n; i++)
      if (h[i] != ti[i])
        gi[ti[i]].insert(i), seet.push_back(i), ed[h[i]].insert(i);
    function<void(int)> dfs = [&](int u) {
      ed[h[u]].erase(u);
      sum++;
      gi[ti[u]].erase(u);
      if (!ed[ti[u]].empty()) dfs(*ed[ti[u]].begin());
    };
    for (int i : seet)
      if (gi[ti[i]].count(i)) {
        sum = 0;
        dfs(i);
        if (sum % 2 == 0) ++ans;
      }
    bool p = 1;
    if (tot > 1 || ans % 2 == 0) p = 0;
    if (!p)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

```