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 Maxn = 1e5 + 1, Mod = 1e9 + 7;
const long long int MOD = 1000000007LL;
const long long int MAX = 100010LL;
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
T power(T x, T y, long long int m = MOD) {
  T ans = 1;
  x %= m;
  while (y > 0) {
    if (y & 1ll) ans = (ans * x) % m;
    y >>= 1ll;
    x = (x * x) % m;
  }
  return ans % m;
}
template <typename T, typename T1>
T mod(T x, T1 p) {
  x %= p;
  if (x < 0) x += p;
  return x;
}
string pos1(int c, int d) {
  if (abs(c - d) > 1) return "";
  string s;
  if (c >= d) {
    while (d--) s += "2 ", s += "3 ", c--;
    while (c--) s += "2 ";
    return s;
  } else {
    while (c--) s += "3 ", s += "2 ", d--;
    while (d--) s += "3 ";
    return s;
  }
}
string pos(int b, int c, int d) {
  string s;
  if (c < b) return "";
  while (b--) s += "1 ", s += "2 ", c--;
  if (d != c && d != c + 1) return "";
  while (c--) s += "3 ", s += "2 ", d--;
  while (d--) s += "3 ";
  return s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  for (int z = 0; z < t; z++) {
    int n;
    cin >> n;
    vector<pair<int, int> > og(n);
    vector<pair<int, int> > arr(n);
    int so = 1, o = 0, ze = 0;
    for (int i = 0; i < n; i++) {
      cin >> arr[i].first;
      if (i > 0 && arr[i].first < arr[i - 1].first) so = 0;
    }
    for (int i = 0; i < n; i++) {
      cin >> arr[i].second;
      if (arr[i].second == 1) o = 1;
      if (arr[i].second == 0) ze = 1;
    }
    og = arr;
    sort(arr.begin(), arr.end());
    if (so) {
      cout << "Yes" << endl;
      continue;
    }
    if (o == 0 || ze == 0) {
      cout << "No" << endl;
      continue;
    }
    cout << "Yes" << endl;
  }
}