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;
int rd() {
  int res = 0, fl = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') fl = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    res = (res << 3) + (res << 1) + c - '0';
    c = getchar();
  }
  return res * fl;
}
int T, n, a[550], b[550], chge[550], fl0, fl1, rise;
bool ok(int i, int j) {
  if (b[i] != b[j]) return 1;
  for (int k = 1; k <= n; ++k) {
    if (a[k] > a[j] && a[k] < a[i] && b[k] != b[i]) return 1;
  }
  return 0;
}
int main() {
  T = rd();
  while (T--) {
    n = rd();
    fl0 = 0, fl1 = 0, rise = 1;
    for (int i = 1; i <= n; ++i) a[i] = rd();
    for (int i = 1; i <= n; ++i) b[i] = rd();
    for (int i = 1; i <= n; ++i) {
      if (b[i] == 0) fl0 = 1;
      if (b[i] == 1) fl1 = 1;
      if (a[i] < a[i - 1]) rise = 0;
    }
    if (fl0 && fl1 || rise)
      printf("Yes\n");
    else
      printf("No\n");
  }
  return 0;
}
