#include <bits/stdc++.h>
using namespace std;
template <typename typC, int N>
struct bit {
  typC a[N], s[N];
  int n;
  bit() {
    memset(s, 0, sizeof s);
    memset(a, 0, sizeof a);
  }
  void init(int nn) {
    n = nn;
    memset(a + 1, 0, n * sizeof a[0]);
    memset(s + 1, 0, n * sizeof a[0]);
  }
  void init(int nn, typC *b) {
    n = nn;
    for (int i = 1; i <= n; i++) s[i] = s[i - 1] + b[i];
    memset(a + 1, 0, n * sizeof a[0]);
  }
  void mdf(int x, typC y) {
    a[x] += y;
    while ((x += x & -x) <= n) a[x] += y;
  }
  typC sum(int x) {
    typC r = a[x];
    while (x ^= x & -x) r += a[x];
    return r;
  }
  typC sum(int x, int y) { return sum(y) - sum(x - 1); }
};
const int N = 5e5 + 2;
bit<int, N> s;
int a[N], ss[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, i;
    long long r = 0;
    cin >> n;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i <= n; i++) ss[i] = 0;
    for (i = 1; i <= n; i++) ss[a[i]]++;
    if (*max_element(ss + 1, ss + n + 1) > 1) {
      cout << "YES\n";
      continue;
    }
    s.init(n);
    for (i = 1; i <= n; i++) r += i - 1 - s.sum(a[i]), s.mdf(a[i], 1);
    if (r & 1)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <typename typC, int N>
struct bit {
  typC a[N], s[N];
  int n;
  bit() {
    memset(s, 0, sizeof s);
    memset(a, 0, sizeof a);
  }
  void init(int nn) {
    n = nn;
    memset(a + 1, 0, n * sizeof a[0]);
    memset(s + 1, 0, n * sizeof a[0]);
  }
  void init(int nn, typC *b) {
    n = nn;
    for (int i = 1; i <= n; i++) s[i] = s[i - 1] + b[i];
    memset(a + 1, 0, n * sizeof a[0]);
  }
  void mdf(int x, typC y) {
    a[x] += y;
    while ((x += x & -x) <= n) a[x] += y;
  }
  typC sum(int x) {
    typC r = a[x];
    while (x ^= x & -x) r += a[x];
    return r;
  }
  typC sum(int x, int y) { return sum(y) - sum(x - 1); }
};
const int N = 5e5 + 2;
bit<int, N> s;
int a[N], ss[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, i;
    long long r = 0;
    cin >> n;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i <= n; i++) ss[i] = 0;
    for (i = 1; i <= n; i++) ss[a[i]]++;
    if (*max_element(ss + 1, ss + n + 1) > 1) {
      cout << "YES\n";
      continue;
    }
    s.init(n);
    for (i = 1; i <= n; i++) r += i - 1 - s.sum(a[i]), s.mdf(a[i], 1);
    if (r & 1)
      cout << "NO\n";
    else
      cout << "YES\n";
  }
}

```