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;
struct node {
  long long nod, d;
  node(long long a, long long b) {
    nod = a;
    d = b;
  }
  bool operator<(const node &p) const { return p.d < d; }
};
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
long long reset(long long n, long long pos) { return n = n & ~(1LL << pos); }
bool check(long long n, long long pos) { return (bool)(n & (1LL << pos)); }
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long n;
long long arr[100005];
long long status[100005];
int main() {
  long long tes;
  scanf("%lld", &tes);
  for (long long cas = 1; cas <= tes; cas++) {
    scanf("%lld", &n);
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &arr[i]);
    }
    bool f1 = 0;
    bool f2 = 0;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &status[i]);
      if (status[i] == 0) f1 = 1;
      if (status[i] == 1) f2 = 1;
    }
    if (f1 && f2) {
      printf("Yes\n");
      continue;
    }
    bool f = 0;
    for (long long i = 1; i <= n - 1; i++) {
      if (arr[i] > arr[i + 1]) {
        f = 1;
        break;
      }
    }
    if (f) {
      printf("No\n");
    } else {
      printf("Yes\n");
    }
  }
  return 0;
}