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) {
  if (x == 0 && y == 0) return 1;
  if (x <= 1 && y != 0) return 0;
  if (y == 0) return 1;
  long long int ans = 1;
  while (y > 0) {
    if (y % 2) ans *= x;
    y = y / 2;
    x = x * x;
  }
  return ans;
}
void solve(vector<pair<int, int>> a, int n) {
  for (int i = 0; i < n - 1; i++) {
    for (int j = 0; j < n - i - 1; j++)
      if (a[j].first > a[j + 1].first) {
        if (a[j].second != a[j + 1].second)
          swap(a[j], a[j + 1]);
        else {
          cout << "No" << endl;
          return;
        }
      }
  }
  cout << "Yes" << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int a[n], b[n];
    int flag = 0;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++) {
      cin >> b[i];
      if (b[0] != b[i]) {
        flag = 1;
      }
    }
    if (flag == 1) {
      cout << "Yes" << endl;
      continue;
    } else {
      int h = 0;
      for (int i = 1; i < n; i++) {
        if (a[i] >= a[i - 1])
          continue;
        else {
          h = 1;
          break;
        }
      }
      if (h == 0)
        cout << "Yes" << endl;
      else {
        cout << "No" << endl;
      }
    }
  }
  return 0;
}