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;
int dp[1000][1000];
long long int mpow(long long int base, long long int exp);
void ipgraph(int m);
void dfs(int u, int par);
const int N = 130, M = N;
vector<long long int> g[N];
int a[N][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long int d = 0, e = 0, p = 0, q = 0, r = 0;
    long long int ans = 0, res = 0, cnt = 0, sum = 0, diff = 0, mul = 1,
                  val = 0;
    string s;
    long long int n, m;
    cin >> n;
    long long int a[n + 1], b[n + 1];
    bool sorted = 1, have0 = 0, have1 = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (i >= 2 && a[i] < a[i - 1]) sorted = 0;
    }
    for (int i = 1; i <= n; i++) {
      cin >> b[i];
      if (!b[i])
        have0 = 1;
      else
        have1 = 1;
    }
    if (have0 && have1)
      cout << "Yes" << '\n';
    else if (sorted)
      cout << "Yes" << '\n';
    else
      cout << "No" << '\n';
  }
  return 0;
}