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 isprime(int n) {
  if (n == 1) return false;
  if (n == 2) return true;
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
void answer(int f) {
  if (f) {
    cout << "YES";
  } else {
    cout << "NO";
  }
  cout << endl;
}
long long int countSetBits(long long int n) {
  long long int count = 0;
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
long long int factorial(long long int n) {
  n = n % 1000000007;
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
long long int FastExp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return (ans % 1000000007);
}
int main() {
  int test = 1;
  cin >> test;
  while (test--) {
    long long int n, i, j;
    cin >> n;
    long long int a[n], b[n], c[n];
    vector<long long int> x, y;
    vector<pair<int, int> > v(n), u(n);
    for (i = 0; i < n; i++) {
      int x;
      cin >> x;
      v[i].first = x;
      u[i].first = x;
    }
    int z = 0, o = 0;
    for (i = 0; i < n; i++) {
      int x;
      cin >> x;
      if (x == 0)
        z++;
      else {
        o++;
      }
      v[i].second = x;
      u[i].second = x;
    }
    int f = 1;
    sort(u.begin(), u.end());
    for (i = 0; i < n; i++) {
      if (v[i].first != u[i].first) f = 0;
    }
    if (f == 0 && (o == 0 || z == 0))
      f = 0;
    else
      f = 1;
    answer(f);
  }
  return 0;
}