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;
void getZarr(string str, vector<long long int> &Z) {
  long long int n = str.length();
  long long int L, R, k;
  L = R = 0;
  for (long long int i = 1; i < n; ++i) {
    if (i > R) {
      L = R = i;
      while (R < n && str[R - L] == str[R]) R++;
      Z[i] = R - L;
      R--;
    } else {
      k = i - L;
      if (Z[k] < R - i + 1)
        Z[i] = Z[k];
      else {
        L = i;
        while (R < n && str[R - L] == str[R]) R++;
        Z[i] = R - L;
        R--;
      }
    }
  }
}
vector<long long int> isprime;
void seive(long long int size) {
  isprime.resize(size + 1, 1);
  long long int i, j;
  isprime[0] = 0;
  isprime[1] = 0;
  for (i = 2; i * i <= size; i++) {
    if (isprime[i]) {
      for (j = 2; j * i <= size; j++) isprime[i * j] = 0;
    }
  }
}
long long int power(long long int x, unsigned long long int y,
                    long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m);
  p = (p * p) % m;
  if (y % 2 != 0) p = (p * x) % m;
  return p;
}
long long int modinverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
long long int findans() {
  long long int n, i, o = 0, z = 0;
  cin >> n;
  vector<long long int> v(n);
  for (i = 0; i < n; i++) cin >> v[i];
  for (i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    if (a == 0)
      z++;
    else
      o++;
  }
  if (z == 0 || o == 0) {
    vector<long long int> v2 = v;
    sort((v2.begin()), (v2.end()));
    for (i = 0; i < n; i++) {
      if (v[i] != v2[i]) {
        cout << "No\n";
        return 0;
      }
    }
    cout << "Yes\n";
    return 0;
  }
  cout << "Yes\n";
  return 0;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t, i;
  cin >> t;
  for (i = 0; i < t; i++) {
    findans();
  }
  return 0;
}