#include <bits/stdc++.h>
using namespace std;
inline void wt(long long x) { cout << x; }
inline void wtl(long long x) { cout << x << '\n'; }
inline void wtb(long long x) { cout << x << ' '; }
inline void wt(char x) { cout << x; }
inline void wtl(char x) { cout << x << '\n'; }
inline void wtb(char x) { cout << x << ' '; }
inline void wt(string x) { cout << x; }
inline void wtl(string x) { cout << x << '\n'; }
inline void wtb(string x) { cout << x << ' '; }
template <typename T>
bool ckmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool ckmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
long long qmi(long long a, long long k, long long p) {
  long long res = 1;
  while (k) {
    if (k & 1) res = (long long)res * a % p;
    a = (long long)a * a % p;
    k >>= 1;
  }
  return res;
}
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    b >>= 1;
    a *= a;
  }
  return res;
}
long long mo(long long x, long long p) { return x = ((x % p) + p) % p; }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const long long maxn = 1e6 + 7;
const long long mod = 1e9 + 7;
long long dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
long long fact[maxn], infact[maxn];
void jie_init(long long n) {
  fact[0] = 1;
  for (long long i = (1); i <= (n); i++) fact[i] = fact[i - 1] * i % mod;
  infact[n] = qmi(fact[n], mod - 2, mod);
  for (long long i = (n - 1); i >= (0); i--)
    infact[i] = infact[i + 1] * (i + 1) % mod;
}
long long C(long long n, long long m) {
  return fact[n] * infact[m] % mod * infact[n - m] % mod;
}
long long T = 1, N, M, K;
long long A[maxn], B[maxn], t[maxn], f[maxn];
long long merge_sort(long long a[], long long l, long long r) {
  if (l == r) return 0;
  long long mid = l + r >> 1;
  long long res = merge_sort(a, l, mid) + merge_sort(a, mid + 1, r);
  long long i = l, j = mid + 1, k = 0;
  while (i <= mid && j <= r) {
    if (a[i] <= a[j])
      t[k++] = a[i++];
    else {
      t[k++] = a[j++];
      res += mid - i + 1;
    }
  }
  while (i <= mid) t[k++] = a[i++];
  while (j <= r) t[k++] = a[j++];
  for (long long i = 0, j = l; j <= r; i++, j++) a[j] = t[i];
  return res;
}
void solve() {
  cin >> N;
  map<long long, long long> st;
  long long ff = 0;
  for (long long i = (1); i <= (N); i++) {
    cin >> A[i];
    if (st[A[i]]) ff = 1;
    st[A[i]] = 1;
  }
  if (ff) {
    puts("YES");
    return;
  }
  long long x = merge_sort(A, 1, N);
  if (x & 1) {
    puts("NO");
  } else
    puts("YES");
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> T;
  while (T--) solve();
  return (0 - 0);
}

### Prompt

Please provide a cpp coded solution 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;
inline void wt(long long x) { cout << x; }
inline void wtl(long long x) { cout << x << '\n'; }
inline void wtb(long long x) { cout << x << ' '; }
inline void wt(char x) { cout << x; }
inline void wtl(char x) { cout << x << '\n'; }
inline void wtb(char x) { cout << x << ' '; }
inline void wt(string x) { cout << x; }
inline void wtl(string x) { cout << x << '\n'; }
inline void wtb(string x) { cout << x << ' '; }
template <typename T>
bool ckmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool ckmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
long long qmi(long long a, long long k, long long p) {
  long long res = 1;
  while (k) {
    if (k & 1) res = (long long)res * a % p;
    a = (long long)a * a % p;
    k >>= 1;
  }
  return res;
}
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    b >>= 1;
    a *= a;
  }
  return res;
}
long long mo(long long x, long long p) { return x = ((x % p) + p) % p; }
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const long long maxn = 1e6 + 7;
const long long mod = 1e9 + 7;
long long dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};
long long fact[maxn], infact[maxn];
void jie_init(long long n) {
  fact[0] = 1;
  for (long long i = (1); i <= (n); i++) fact[i] = fact[i - 1] * i % mod;
  infact[n] = qmi(fact[n], mod - 2, mod);
  for (long long i = (n - 1); i >= (0); i--)
    infact[i] = infact[i + 1] * (i + 1) % mod;
}
long long C(long long n, long long m) {
  return fact[n] * infact[m] % mod * infact[n - m] % mod;
}
long long T = 1, N, M, K;
long long A[maxn], B[maxn], t[maxn], f[maxn];
long long merge_sort(long long a[], long long l, long long r) {
  if (l == r) return 0;
  long long mid = l + r >> 1;
  long long res = merge_sort(a, l, mid) + merge_sort(a, mid + 1, r);
  long long i = l, j = mid + 1, k = 0;
  while (i <= mid && j <= r) {
    if (a[i] <= a[j])
      t[k++] = a[i++];
    else {
      t[k++] = a[j++];
      res += mid - i + 1;
    }
  }
  while (i <= mid) t[k++] = a[i++];
  while (j <= r) t[k++] = a[j++];
  for (long long i = 0, j = l; j <= r; i++, j++) a[j] = t[i];
  return res;
}
void solve() {
  cin >> N;
  map<long long, long long> st;
  long long ff = 0;
  for (long long i = (1); i <= (N); i++) {
    cin >> A[i];
    if (st[A[i]]) ff = 1;
    st[A[i]] = 1;
  }
  if (ff) {
    puts("YES");
    return;
  }
  long long x = merge_sort(A, 1, N);
  if (x & 1) {
    puts("NO");
  } else
    puts("YES");
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> T;
  while (T--) solve();
  return (0 - 0);
}

```