#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000009;
vector<int> seg;
int seg_q(int num, int le, int ri, int x, int y) {
  if (y < le || ri < x)
    return 0;
  else if (x <= le && ri <= y) {
    return seg[num];
  }
  int mid = (le + ri) / 2;
  return seg_q(num * 2, le, mid, x, y) + seg_q(num * 2 + 1, mid + 1, ri, x, y);
}
int seg_u(int num, int le, int ri, int x) {
  if (x < le || ri < x)
    return seg[num];
  else if (x == le && le == ri)
    return ++seg[num];
  int mid = (le + ri) / 2;
  return seg[num] =
             seg_u(num * 2, le, mid, x) + seg_u(num * 2 + 1, mid + 1, ri, x);
}
void init() {}
void solve() {
  int n;
  cin >> n;
  seg = vector<int>(n * 4 + 4);
  long long cnt = 0;
  bool ch = false;
  for (int i = 0; i < (int)(n); ++i) {
    int x;
    cin >> x;
    cnt += seg_q(1, 1, n, x + 1, n);
    if (!ch && seg_q(1, 1, n, x, x) > 0) ch = true;
    seg_u(1, 1, n, x);
  }
  cout << (cnt % 2 && !ch ? "NO" : "YES") << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  init();
  int test_cases = 1;
  cin >> test_cases;
  while (test_cases--) {
    solve();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 mod = 1000000009;
vector<int> seg;
int seg_q(int num, int le, int ri, int x, int y) {
  if (y < le || ri < x)
    return 0;
  else if (x <= le && ri <= y) {
    return seg[num];
  }
  int mid = (le + ri) / 2;
  return seg_q(num * 2, le, mid, x, y) + seg_q(num * 2 + 1, mid + 1, ri, x, y);
}
int seg_u(int num, int le, int ri, int x) {
  if (x < le || ri < x)
    return seg[num];
  else if (x == le && le == ri)
    return ++seg[num];
  int mid = (le + ri) / 2;
  return seg[num] =
             seg_u(num * 2, le, mid, x) + seg_u(num * 2 + 1, mid + 1, ri, x);
}
void init() {}
void solve() {
  int n;
  cin >> n;
  seg = vector<int>(n * 4 + 4);
  long long cnt = 0;
  bool ch = false;
  for (int i = 0; i < (int)(n); ++i) {
    int x;
    cin >> x;
    cnt += seg_q(1, 1, n, x + 1, n);
    if (!ch && seg_q(1, 1, n, x, x) > 0) ch = true;
    seg_u(1, 1, n, x);
  }
  cout << (cnt % 2 && !ch ? "NO" : "YES") << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  init();
  int test_cases = 1;
  cin >> test_cases;
  while (test_cases--) {
    solve();
  }
  return 0;
}

```