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;
long long int power(long long int x, long long int y, long long int m) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long int nCr(long long int n, long long int r, long long int m) {
  if (r > n) return 0;
  long long int a = 1, b = 1, i;
  for (i = 0; i < r; i++) {
    a = (a * n) % m;
    --n;
  }
  while (r) {
    b = (b * r) % m;
    --r;
  }
  return (a * power(b, m - 2, m)) % m;
}
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n), brr(n);
  for (int i = 0; i < n; ++i) cin >> arr[i];
  for (int i = 0; i < n; ++i) cin >> brr[i];
  int mx0 = 0, mx1 = 0, f = 0;
  for (int i = 0; i < n - 1; ++i) {
    if (arr[i] > arr[i + 1]) f = 1;
  }
  int f2 = 0;
  for (int i = 0; i < n - 1; ++i) {
    if (brr[i] != brr[i + 1]) f2 = 1;
  }
  if (!f || f2) {
    cout << "Yes" << '\n';
  } else
    cout << "No" << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}