Problem: Ashish has n elements arranged in a line. 

These elements are represented by two integers a_i — the value of the element and b_i — the type of the element (there are only two possible types: 0 and 1). He wants to sort the elements in non-decreasing values of a_i.

He can perform the following operation any number of times: 

  * Select any two elements i and j such that b_i ≠ b_j and swap them. That is, he can only swap two elements of different types in one move. 



Tell him if he can sort the elements in non-decreasing values of a_i after performing any number of operations.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains one integer n (1 ≤ n ≤ 500) — the size of the arrays.

The second line contains n integers a_i (1 ≤ a_i ≤ 10^5) — the value of the i-th element.

The third line containts n integers b_i (b_i ∈ \{0, 1\}) — the type of the i-th element.

Output

For each test case, print "Yes" or "No" (without quotes) depending on whether it is possible to sort elements in non-decreasing order of their value.

You may print each letter in any case (upper or lower).

Example

Input


5
4
10 20 20 30
0 1 0 1
3
3 1 2
0 1 1
4
2 2 4 8
1 1 1 1
3
5 15 4
0 0 0
4
20 10 100 50
1 0 0 1


Output


Yes
Yes
Yes
No
Yes

Note

For the first case: The elements are already in sorted order.

For the second case: Ashish may first swap elements at positions 1 and 2, then swap elements at positions 2 and 3.

For the third case: The elements are already in sorted order.

For the fourth case: No swap operations may be performed as there is no pair of elements i and j such that b_i ≠ b_j. The elements cannot be sorted.

For the fifth case: Ashish may swap elements at positions 3 and 4, then elements at positions 1 and 2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  cout << "x= " << x << endl;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  cout << "res" << res << endl;
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p) % p;
}
long long int nCrModPFermat(long long int n, long long int r, long long int p) {
  if (r == 0) return 1;
  long long int fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n], b[n];
    int zeroes = 0, ones = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      b[i] = a[i];
    }
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      if (x)
        ones++;
      else
        zeroes++;
    }
    bool f = 1;
    sort(a, a + n);
    if (ones == n || zeroes == n) {
      for (int i = 0; i < n; i++) {
        if (a[i] != b[i]) {
          cout << "No" << endl;
          f = 0;
          break;
        }
      }
      if (f) {
        cout << "Yes\n";
      }
    } else {
      cout << "Yes\n";
    }
  }
}