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.
#include <bits/stdc++.h>
using namespace std;
const long long MAX_VALUE = LONG_MAX;
const long long MIN_VALUE = LONG_MIN;
const unsigned long long U_MAX_VALUE = ULLONG_MAX;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n], t[n], b[n], zero_count = 0, one_count = 0, need_to_sort = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    t[i] = a[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
    if (b[i] == 0)
      zero_count++;
    else
      one_count++;
  }
  sort(t, t + n);
  for (long long i = 0; i < n; i++) {
    if (a[i] != t[i]) {
      need_to_sort++;
    }
  }
  if (need_to_sort == 0 || (one_count > 0 && zero_count > 0)) {
    cout << "Yes";
    return;
  }
  cout << "No";
  return;
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long t = 1, y = 2;
  cin >> t;
  while (t--) {
    solve();
    cout << "\n";
  }
}
