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>
const long long N = 1 << 22;
using namespace std;
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  return b ? gcd(b, a % b) : a;
}
unsigned long long lcm(unsigned long long a, unsigned long long b) {
  return a * b / gcd(a, b);
}
template <class T>
T MIN(const T& a, const T& b) {
  return a < b ? a : b;
}
template <class T>
T MAX(const T& a, const T& b) {
  return a > b ? a : b;
}
long long bit[N + 1];
void update(long long pos, long long val) {
  while (pos <= N) {
    bit[pos] += val;
    pos += pos & (-pos);
  }
}
long long bsearch(long long x) {
  if (x == 0) return 0;
  long long pos = 0;
  for (long long i = N / 2; i >= 1; i >>= 1) {
    if (bit[i + pos] < x) {
      x -= bit[i + pos];
      pos += i;
    }
  }
  return pos + 1;
}
long long getsum(long long pos) {
  long long sum = 0;
  while (pos > 0) {
    sum += bit[pos];
    pos -= pos & (-pos);
  }
  return sum;
}
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    vector<long long> b(n);
    for (long long i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long f = 0;
    for (long long i = 0; i < n; i++) {
      cin >> b[i];
      if (!f) {
        if (b[i])
          f = 1;
        else
          f = -1;
      } else if (f == 1 && !b[i])
        f = 2;
      else if (f == -1 && b[i])
        f = 2;
    }
    if (f == 2) {
      cout << "Yes" << endl;
      continue;
    }
    f = 0;
    for (long long i = 1; i < n; i++) {
      if (a[i] < a[i - 1]) f = 1;
    }
    if (f)
      cout << "No" << endl;
    else
      cout << "Yes" << endl;
  }
  return 0;
}