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;
const int N = 1e4 + 100;
const int M = 1e6 + 2;
const long long mod = 998244353;
const int inf = 2e9 + 3;
const long long INF = 1e18;
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
void data() {}
int T, n, a[N], b[N];
int main() {
  data();
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for (int i = 1; i <= n; ++i) scanf("%d", &b[i]);
    vector<int> v0, v1, ans;
    bool f = 0;
    for (int i = 1; i <= n; ++i) {
      if (b[i] == 0) v0.push_back(a[i]);
      if (b[i] == 1) v1.push_back(a[i]);
    }
    if ((int)v0.size() && (int)v1.size()) {
      printf("Yes\n");
      continue;
    }
    reverse(v0.begin(), v0.end());
    reverse(v1.begin(), v1.end());
    for (int i = 1; i <= n; ++i) {
      if (v1.empty()) {
        ans.push_back(v0.back());
        v0.pop_back();
        continue;
      }
      if (v0.empty()) {
        ans.push_back(v1.back());
        v1.pop_back();
        continue;
      }
      if (v1.back() < v0.back()) {
        ans.push_back(v1.back());
        v1.pop_back();
      } else {
        ans.push_back(v0.back());
        v0.pop_back();
      }
    }
    for (int i = 1; i < (int)ans.size(); ++i) {
      if (ans[i] < ans[i - 1]) f = 1;
    }
    if (!f)
      printf("Yes\n");
    else
      printf("No\n");
  }
}