#include <bits/stdc++.h>
#pragma GCC optimize "trapv"
using namespace std;
template <typename T1, typename T2>
istream &operator>>(istream &istream, pair<T1, T2> &p) {
  return (istream >> p.first >> p.second);
}
template <typename T>
istream &operator>>(istream &istream, vector<T> &v) {
  for (auto &it : v) cin >> it;
  return istream;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
  return (ostream << p.first << " " << p.second);
}
template <typename T>
ostream &operator<<(ostream &ostream, const vector<T> &c) {
  for (auto &it : c) cout << it << " ";
  return ostream;
}
vector<pair<int, int> > dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
struct DSU {
  vector<int> par;
  vector<int> sze;
  int connected;
  void ini(int n) {
    par.resize(n + 1);
    sze.assign(n + 1, 1);
    iota((par).begin(), (par).end(), 0);
    connected = n;
  }
  int get(int x) { return (par[x] = (par[x] == x ? x : get(par[x]))); }
  int getSize(int u) { return sze[get(u)]; }
  void un(int x, int y) {
    x = get(x);
    y = get(y);
    if (x == y) {
      return;
    }
    sze[y] += sze[x];
    sze[x] = 0;
    par[x] = y;
    connected--;
  }
};
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  cin >> v;
  vector<int> vc = v;
  sort((vc).begin(), (vc).end());
  map<int, int> m;
  for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) m[v[i]]++;
  if (m.size() < n) {
    cout << "YES\n";
    return;
  }
  DSU dsu;
  dsu.ini(n + 1);
  for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1)
    dsu.un(v[i], i + 1);
  int s = 0;
  map<int, int> f;
  for (int i = 1; 1 < n + 1 ? i < n + 1 : i > n + 1;
       1 < n + 1 ? i += 1 : i -= 1) {
    if (f[dsu.get(i)] == 0 && dsu.getSize(i) % 2 == 0) {
      s++;
      f[dsu.get(i)] = 1;
    }
  }
  if (s % 2)
    cout << "NO";
  else
    cout << "YES";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cout << fixed << setprecision(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  };
  return 0;
}

### Prompt

Generate 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>
#pragma GCC optimize "trapv"
using namespace std;
template <typename T1, typename T2>
istream &operator>>(istream &istream, pair<T1, T2> &p) {
  return (istream >> p.first >> p.second);
}
template <typename T>
istream &operator>>(istream &istream, vector<T> &v) {
  for (auto &it : v) cin >> it;
  return istream;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
  return (ostream << p.first << " " << p.second);
}
template <typename T>
ostream &operator<<(ostream &ostream, const vector<T> &c) {
  for (auto &it : c) cout << it << " ";
  return ostream;
}
vector<pair<int, int> > dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
struct DSU {
  vector<int> par;
  vector<int> sze;
  int connected;
  void ini(int n) {
    par.resize(n + 1);
    sze.assign(n + 1, 1);
    iota((par).begin(), (par).end(), 0);
    connected = n;
  }
  int get(int x) { return (par[x] = (par[x] == x ? x : get(par[x]))); }
  int getSize(int u) { return sze[get(u)]; }
  void un(int x, int y) {
    x = get(x);
    y = get(y);
    if (x == y) {
      return;
    }
    sze[y] += sze[x];
    sze[x] = 0;
    par[x] = y;
    connected--;
  }
};
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  cin >> v;
  vector<int> vc = v;
  sort((vc).begin(), (vc).end());
  map<int, int> m;
  for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1) m[v[i]]++;
  if (m.size() < n) {
    cout << "YES\n";
    return;
  }
  DSU dsu;
  dsu.ini(n + 1);
  for (int i = 0; 0 < n ? i < n : i > n; 0 < n ? i += 1 : i -= 1)
    dsu.un(v[i], i + 1);
  int s = 0;
  map<int, int> f;
  for (int i = 1; 1 < n + 1 ? i < n + 1 : i > n + 1;
       1 < n + 1 ? i += 1 : i -= 1) {
    if (f[dsu.get(i)] == 0 && dsu.getSize(i) % 2 == 0) {
      s++;
      f[dsu.get(i)] = 1;
    }
  }
  if (s % 2)
    cout << "NO";
  else
    cout << "YES";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cout << fixed << setprecision(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  };
  return 0;
}

```