#include <bits/stdc++.h>
using namespace std;
vector<int> tree;
int get(int v, int l, int r, int ql, int qr) {
  if (l >= qr || ql >= r) return 0;
  if (ql <= l && r <= qr) return tree[v];
  return get(2 * v + 1, l, (l + r) / 2, ql, qr) +
         get(2 * v + 2, (l + r) / 2, r, ql, qr);
}
void upd(int v, int l, int r, int ql, int qr) {
  if (l >= qr || ql >= r) return;
  if (ql <= l && r <= qr) {
    tree[v]++;
    return;
  }
  upd(2 * v + 1, l, (l + r) / 2, ql, qr);
  upd(2 * v + 2, (l + r) / 2, r, ql, qr);
  tree[v] = tree[2 * v + 1] + tree[2 * v + 2];
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<bool> used(n);
  bool check = false;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
    if (used[a[i]]) check = true;
    used[a[i]] = true;
  }
  if (check) {
    cout << "YES\n";
    return;
  }
  tree.assign(4 * n, 0);
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] < n - 1) ans += 1ll * get(0, 0, n, a[i] + 1, n);
    upd(0, 0, n, a[i], a[i] + 1);
  }
  if (ans % 2 == 0)
    cout << "YES\n";
  else
    cout << "NO\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
vector<int> tree;
int get(int v, int l, int r, int ql, int qr) {
  if (l >= qr || ql >= r) return 0;
  if (ql <= l && r <= qr) return tree[v];
  return get(2 * v + 1, l, (l + r) / 2, ql, qr) +
         get(2 * v + 2, (l + r) / 2, r, ql, qr);
}
void upd(int v, int l, int r, int ql, int qr) {
  if (l >= qr || ql >= r) return;
  if (ql <= l && r <= qr) {
    tree[v]++;
    return;
  }
  upd(2 * v + 1, l, (l + r) / 2, ql, qr);
  upd(2 * v + 2, (l + r) / 2, r, ql, qr);
  tree[v] = tree[2 * v + 1] + tree[2 * v + 2];
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<bool> used(n);
  bool check = false;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    a[i]--;
    if (used[a[i]]) check = true;
    used[a[i]] = true;
  }
  if (check) {
    cout << "YES\n";
    return;
  }
  tree.assign(4 * n, 0);
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    if (a[i] < n - 1) ans += 1ll * get(0, 0, n, a[i] + 1, n);
    upd(0, 0, n, a[i], a[i] + 1);
  }
  if (ans % 2 == 0)
    cout << "YES\n";
  else
    cout << "NO\n";
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```