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>
char *fs, *ft, buf[1 << 20];
inline int read() {
  int x = 0, f = 1;
  char ch =
      (fs == ft && (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
          ? 0
          : *fs++;
  ;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
             ? 0
             : *fs++;
    ;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 20, stdin), fs == ft))
             ? 0
             : *fs++;
    ;
  }
  return x * f;
}
using namespace std;
int a[1009];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    int z0 = 0, z1 = 0, bo = 0;
    cin >> n;
    cin >> a[1];
    for (int i = 2; i <= n; i++) {
      cin >> a[i];
      if (a[i] < a[i - 1]) bo = 1;
    }
    for (int i = 1; i <= n; i++) {
      int k;
      cin >> k;
      if (k == 0)
        z0++;
      else
        z1++;
    }
    if (bo == 0) {
      cout << "Yes" << endl;
      continue;
    } else {
      if (z0 == 0 || z1 == 0)
        cout << "No" << endl;
      else
        cout << "Yes" << endl;
    }
  }
  return 0;
}