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;
const int CONST = 1e9 + 7;
const double pi = 3.14159265358979323846;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a, b;
    for (int i = 0; i < n; ++i) {
      int tmp;
      cin >> tmp;
      a.push_back(tmp);
    }
    int sum = 0;
    for (int i = 0; i < n; ++i) {
      int tmp;
      cin >> tmp;
      b.push_back(tmp);
      sum += tmp;
    }
    bool f = false;
    if (sum == n || sum == 0) {
      for (int i = 1; i < n; ++i) {
        if (a[i] < a[i - 1]) {
          f = true;
          break;
        }
      }
      if (!f)
        cout << "Yes" << endl;
      else
        cout << "No" << endl;
    } else {
      cout << "Yes" << endl;
    }
  }
  return 0;
}