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 INF = 1e9;
const int MOD = 1e9 + 7;
void _debug() { cerr << "]" << endl; }
template <typename T>
void _debug(T a) {
  cerr << a << ']' << endl;
}
template <typename T, typename... Args>
void _debug(T a, Args... args) {
  cerr << a << ", ";
  _debug(args...);
}
template <typename T>
void read(T &a) {
  cin >> a;
}
template <typename T, typename... Args>
void read(T &a, Args &...args) {
  cin >> a;
  read(args...);
}
template <typename T>
void print(T &&a) {
  cout << a;
}
template <typename T, typename... Args>
void print(T &&a, Args &&...args) {
  cout << a;
  print(forward<Args>(args)...);
}
vector<string> split(string_view s, char delim = ' ') {
  vector<string> res;
  string cur;
  for (int i = 0; i < (int)((s).size()); ++i) {
    if (s[i] == delim) {
      if (!cur.empty()) {
        res.push_back(cur);
        cur = "";
      }
    } else {
      cur += s[i];
    }
  }
  if (!cur.empty()) {
    res.push_back(cur);
  }
  return res;
}
bool is_prime(int a) {
  for (int i = 2; i * i <= a; ++i) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
template <typename T, typename Comp = less<T>>
vector<int> arg_sort(vector<T> &v, Comp comp = Comp()) {
  vector<int> res(v.size());
  iota((res).begin(), (res).end(), 0);
  sort((res).begin(), (res).end(),
       [&](int a, int b) { return comp(v[a], v[b]); });
  return res;
}
template <typename T>
bool in(T a, T x, T y) {
  return x <= a && a < y;
}
int d4i[4] = {0, 1, 0, -1}, d4j[4] = {1, 0, -1, 0};
int d8i[8] = {1, 1, 1, 0, -1, -1, -1, 0}, d8j[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int n;
void solve() {
  int n;
  read(n);
  vector<int> a(n), b(n);
  for (int i = 0; i < (n); ++i) read(a[i]);
  for (int i = 0; i < (n); ++i) read(b[i]);
  if (is_sorted((a).begin(), (a).end())) {
    print("Yes\n");
  } else if (all_of((b).begin(), (b).end(), [&](int i) { return i == b[0]; })) {
    print("No\n");
  } else {
    print("Yes\n");
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed << setprecision(7);
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) solve();
}