#include <bits/stdc++.h>
using namespace std;
void swap(int &x, int &y) {
  int temp = x;
  x = y;
  y = temp;
}
int XOR(int n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
int isPrime(int N) {
  if (N < 2 || (!(N & 1) && N != 2)) return 0;
  for (int i = 3; i * i <= N; i += 2) {
    if (!(N % i)) return 0;
  }
  return 1;
}
void fillPrimes(vector<int> &prime, int high) {
  bool ck[high + 1];
  memset(ck, true, sizeof(ck));
  ck[1] = false;
  ck[0] = false;
  for (int i = 2; (i * i) <= high; i++) {
    if (ck[i] == true) {
      for (int j = i * i; j <= high; j = j + i) {
        ck[j] = false;
      }
    }
  }
  for (int i = 2; i * i <= high; i++) {
    if (ck[i] == true) {
      prime.push_back(i);
    }
  }
}
vector<bool> segmentedSieve(int low, int high) {
  vector<bool> prime(high - low + 1, true);
  vector<int> chprime;
  fillPrimes(chprime, high);
  for (int i : chprime) {
    int lower = (low / i);
    if (lower <= 1) {
      lower = i + i;
    } else if (low % i) {
      lower = (lower * i) + i;
    } else {
      lower = (lower * i);
    }
    for (int j = lower; j <= high; j = j + i) {
      prime[j - low] = false;
    }
  }
  return prime;
}
const double eps = 1e-7;
const long long N = 5e5 + 7, mod = 1e9 + 7;
long long POW(long long a, long long b, long long mod) {
  a %= mod;
  long long r = 1;
  for (long long i = b; i > 0; i >>= 1) {
    if (i & 1) r = (r * a) % mod;
    a = (a * a) % mod;
  }
  return r;
}
long long f[N];
void init() {
  f[0] = 1;
  for (long long i = 1; i < N; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
}
long long nCr(long long n, long long r) {
  if (n < r) return 0;
  return f[n] * POW(f[n - r] * f[r], mod - 2, mod) % mod;
}
long long nPr(long long n, long long r) { return nCr(n, r) * f[r] % mod; }
long long p = 1e9 + 7;
long long modadd(long long a, long long b) {
  a = a % p;
  b = b % p;
  return (((a + b) % p) + p) % p;
}
long long modmul(long long a, long long b) {
  a = a % p;
  b = b % p;
  return (((a * b) % p) + p) % p;
}
long long modsub(long long a, long long b) {
  a = a % p;
  b = b % p;
  return (((a - b) % p) + p) % p;
}
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  for (int i = 0; i < n; i += 1) cin >> arr[i];
  map<long long, long long> m;
  long long f = 0;
  for (int i = 0; i < n; i += 1) {
    m[arr[i]]++;
    if (m[arr[i]] > 1) {
      f = 1;
      break;
    }
  }
  if (f == 1) {
    cout << "YES" << endl;
    return;
  }
  bool vis[n + 1];
  fill(vis, vis + n, 0);
  long long c = 0;
  for (int i = 0; i < n; i += 1) {
    if (!vis[i]) {
      long long s = 1;
      vis[i] = 1;
      long long t = arr[i];
      while (1) {
        if (!vis[t - 1]) {
          s++;
          vis[t - 1] = 1;
          t = arr[t - 1];
        } else
          break;
      }
      if (s % 2 == 0) c++;
    }
  }
  if (c % 2 == 0)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
void swap(int &x, int &y) {
  int temp = x;
  x = y;
  y = temp;
}
int XOR(int n) {
  if (n % 4 == 0) return n;
  if (n % 4 == 1) return 1;
  if (n % 4 == 2) return n + 1;
  return 0;
}
int isPrime(int N) {
  if (N < 2 || (!(N & 1) && N != 2)) return 0;
  for (int i = 3; i * i <= N; i += 2) {
    if (!(N % i)) return 0;
  }
  return 1;
}
void fillPrimes(vector<int> &prime, int high) {
  bool ck[high + 1];
  memset(ck, true, sizeof(ck));
  ck[1] = false;
  ck[0] = false;
  for (int i = 2; (i * i) <= high; i++) {
    if (ck[i] == true) {
      for (int j = i * i; j <= high; j = j + i) {
        ck[j] = false;
      }
    }
  }
  for (int i = 2; i * i <= high; i++) {
    if (ck[i] == true) {
      prime.push_back(i);
    }
  }
}
vector<bool> segmentedSieve(int low, int high) {
  vector<bool> prime(high - low + 1, true);
  vector<int> chprime;
  fillPrimes(chprime, high);
  for (int i : chprime) {
    int lower = (low / i);
    if (lower <= 1) {
      lower = i + i;
    } else if (low % i) {
      lower = (lower * i) + i;
    } else {
      lower = (lower * i);
    }
    for (int j = lower; j <= high; j = j + i) {
      prime[j - low] = false;
    }
  }
  return prime;
}
const double eps = 1e-7;
const long long N = 5e5 + 7, mod = 1e9 + 7;
long long POW(long long a, long long b, long long mod) {
  a %= mod;
  long long r = 1;
  for (long long i = b; i > 0; i >>= 1) {
    if (i & 1) r = (r * a) % mod;
    a = (a * a) % mod;
  }
  return r;
}
long long f[N];
void init() {
  f[0] = 1;
  for (long long i = 1; i < N; i++) {
    f[i] = (f[i - 1] * i) % mod;
  }
}
long long nCr(long long n, long long r) {
  if (n < r) return 0;
  return f[n] * POW(f[n - r] * f[r], mod - 2, mod) % mod;
}
long long nPr(long long n, long long r) { return nCr(n, r) * f[r] % mod; }
long long p = 1e9 + 7;
long long modadd(long long a, long long b) {
  a = a % p;
  b = b % p;
  return (((a + b) % p) + p) % p;
}
long long modmul(long long a, long long b) {
  a = a % p;
  b = b % p;
  return (((a * b) % p) + p) % p;
}
long long modsub(long long a, long long b) {
  a = a % p;
  b = b % p;
  return (((a - b) % p) + p) % p;
}
void solve() {
  long long n;
  cin >> n;
  long long arr[n];
  for (int i = 0; i < n; i += 1) cin >> arr[i];
  map<long long, long long> m;
  long long f = 0;
  for (int i = 0; i < n; i += 1) {
    m[arr[i]]++;
    if (m[arr[i]] > 1) {
      f = 1;
      break;
    }
  }
  if (f == 1) {
    cout << "YES" << endl;
    return;
  }
  bool vis[n + 1];
  fill(vis, vis + n, 0);
  long long c = 0;
  for (int i = 0; i < n; i += 1) {
    if (!vis[i]) {
      long long s = 1;
      vis[i] = 1;
      long long t = arr[i];
      while (1) {
        if (!vis[t - 1]) {
          s++;
          vis[t - 1] = 1;
          t = arr[t - 1];
        } else
          break;
      }
      if (s % 2 == 0) c++;
    }
  }
  if (c % 2 == 0)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```