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;
bool prime[1000000];
void SieveOfEratosthenes(long long int n) {
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
bool is_sort(long long int arr[], long long int n) {
  for (long long int i = 0; i < n - 1; i++) {
    if (arr[i] > arr[i + 1]) return false;
  }
  return true;
}
void solve() {
  long long int n;
  cin >> n;
  long long int ar[n], br[n];
  long long int count_0 = 0;
  long long int count_1 = 0;
  for (long long int i = 0; i < n; i++) cin >> ar[i];
  for (long long int i = 0; i < n; i++) {
    cin >> br[i];
    if (br[i] == 0)
      count_0++;
    else
      count_1++;
  }
  if (is_sort(ar, n)) {
    cout << "Yes";
    return;
  }
  if (count_1 && count_0)
    cout << "Yes";
  else
    cout << "No";
}
signed main() {
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}