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 gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
int main() {
  ios_base::sync_with_stdio(false);
  long long int t;
  cin >> t;
  for (long long int i = 0; i < t; i++) {
    long long int n;
    cin >> n;
    long long int a[2][n];
    for (long long int j = 0; j < n; j++) {
      cin >> a[0][j];
    }
    for (long long int j = 0; j < n; j++) {
      cin >> a[1][j];
    }
    long long int z = 0, o = 0, b;
    for (long long int j = 0; j < n; j++) {
      if (a[1][j] == 0) {
        z = 1;
        break;
      }
    }
    for (long long int j = 0; j < n; j++) {
      if (a[1][j] == 1) {
        o = 1;
        break;
      }
    }
    b = z * o;
    if (b) {
      cout << "YES" << endl;
      continue;
    } else {
      b = 1;
      for (long long int k = 0; k < n - 1; k++) {
        if (a[0][k] > a[0][k + 1]) {
          b = 0;
        }
      }
    }
    if (b) {
      cout << "YES" << endl;
      continue;
    } else {
      cout << "NO" << endl;
      continue;
    }
  }
}