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;
long long int power(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
long long int modInverse(long long int a) { return power(a, 1000000007 - 2); }
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
const int N = 500023;
bool vis[N];
vector<int> adj[N];
void solve() {
  int n;
  cin >> n;
  vector<int> a(n), b(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  for (long long int i = 0; i < n; i++) cin >> b[i];
  int sorted = 1;
  for (int i = 1; i < n; i++) {
    if (a[i] >= a[i - 1]) {
      continue;
    } else {
      sorted = 0;
      break;
    }
  }
  if (sorted == 1) {
    cout << "Yes" << endl;
    return;
  }
  int type0 = 0, type1 = 0;
  for (long long int i = 0; i < n; i++) {
    if (b[i] == 0) type0 = 1;
    if (b[i] == 1) type1 = 1;
  }
  if (type1 && type0) {
    cout << "Yes" << endl;
  } else {
    cout << "No" << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  cin >> T;
  int t = 0;
  while (t++ < T) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}