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 long long INF = 1e17;
const long long MOD = 1000000007;
const long long MAX = 4000001;
const double eps = 1E-18;
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  }
  return b;
}
long long min(long long a, long long b) {
  if (a > b) {
    return b;
  }
  return a;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  if (a < b) {
    return gcd(b, a);
  }
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
struct edge {
  long long ind;
  long long fr;
  long long to;
  long long d;
};
class mint {
  long long x;

 public:
  mint(long long x = 0) : x((x % MOD + MOD) % MOD) {}
  mint operator-() const { return mint(-x); }
  mint& operator+=(const mint& a) {
    if ((x += a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator-=(const mint& a) {
    if ((x += MOD - a.x) >= MOD) x -= MOD;
    return *this;
  }
  mint& operator*=(const mint& a) {
    (x *= a.x) %= MOD;
    return *this;
  }
  mint operator+(const mint& a) const {
    mint res(*this);
    return res += a;
  }
  mint operator-(const mint& a) const {
    mint res(*this);
    return res -= a;
  }
  mint operator*(const mint& a) const {
    mint res(*this);
    return res *= a;
  }
  mint pow(long long t) const {
    if (!t) return 1;
    mint a = pow(t >> 1);
    a *= a;
    if (t & 1) a *= *this;
    return a;
  }
  mint inv() const { return pow(MOD - 2); }
  mint& operator/=(const mint& a) { return (*this) *= a.inv(); }
  mint operator/(const mint& a) const {
    mint res(*this);
    return res /= a;
  }
  friend ostream& operator<<(ostream& os, const mint& m) {
    os << m.x;
    return os;
  }
};
mint pw(mint a, long long b) {
  if (b == 0) {
    return 1;
  }
  mint ret = pw(a, b >> 1);
  ret *= ret;
  if (b & 1) {
    ret *= a;
  }
  return ret;
}
int main() {
  long long T;
  cin >> T;
  for (long long t = 1; t <= T; t++) {
    long long N;
    cin >> N;
    vector<long long> a(N), b(N);
    for (long long i = 0; i < N; i++) cin >> a[i];
    for (long long i = 0; i < N; i++) cin >> b[i];
    long long c[2] = {};
    for (long long i = 0; i < N; i++) c[b[i]]++;
    string ans = "Yes";
    if (c[0] == 0 || c[1] == 0) {
      for (long long i = 0; i < N - 1; i++) {
        if (a[i] > a[i + 1]) {
          ans = "No";
        }
      }
    }
    cout << ans << endl;
  }
}