You are given a colored permutation p_1, p_2, ..., p_n. The i-th element of the permutation has color c_i.

Let's define an infinite path as infinite sequence i, p[i], p[p[i]], p[p[p[i]]] ... where all elements have same color (c[i] = c[p[i]] = c[p[p[i]]] = ...).

We can also define a multiplication of permutations a and b as permutation c = a × b where c[i] = b[a[i]]. Moreover, we can define a power k of permutation p as p^k=\underbrace{p × p × ... × p}_{k  times}.

Find the minimum k > 0 such that p^k has at least one infinite path (i.e. there is a position i in p^k such that the sequence starting from i is an infinite path).

It can be proved that the answer always exists.

Input

The first line contains single integer T (1 ≤ T ≤ 10^4) — the number of test cases.

Next 3T lines contain test cases — one per three lines. The first line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the size of the permutation.

The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n, p_i ≠ p_j for i ≠ j) — the permutation p.

The third line contains n integers c_1, c_2, ..., c_n (1 ≤ c_i ≤ n) — the colors of elements of the permutation.

It is guaranteed that the total sum of n doesn't exceed 2 ⋅ 10^5.

Output

Print T integers — one per test case. For each test case print minimum k > 0 such that p^k has at least one infinite path.

Example

Input


3
4
1 3 4 2
1 2 2 3
5
2 3 4 5 1
1 2 3 4 5
8
7 4 5 6 1 8 3 2
5 3 6 4 7 5 8 4


Output


1
5
2

Note

In the first test case, p^1 = p = [1, 3, 4, 2] and the sequence starting from 1: 1, p[1] = 1, ... is an infinite path.

In the second test case, p^5 = [1, 2, 3, 4, 5] and it obviously contains several infinite paths.

In the third test case, p^2 = [3, 6, 1, 8, 7, 2, 5, 4] and the sequence starting from 4: 4, p^2[4]=8, p^2[8]=4, ... is an infinite path since c_4 = c_8 = 4.
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
vector<int> divisors(int n) {
  vector<int> div;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        div.push_back(i);
      } else {
        div.push_back(i);
        div.push_back(n / i);
      }
    }
  }
  return div;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> p(n), c(n);
    for (int i = 0; i < n; i++) {
      cin >> p[i];
      p[i]--;
    }
    int col;
    for (int i = 0; i < n; i++) {
      cin >> col;
      c[p[i]] = col;
    }
    vector<vector<int>> cycles;
    vector<int> cycle;
    vector<bool> visit(n, false);
    for (int i = 0; i < n; i++) {
      int s = p[i];
      if (visit[s]) {
        continue;
      }
      cycle.clear();
      while (!visit[s]) {
        cycle.push_back(s);
        visit[s] = true;
        s = p[s];
      }
      cycles.push_back(cycle);
    }
    int _min = INT_MAX;
    for (auto cycle : cycles) {
      int len = cycle.size();
      for (int val = 1; val <= len; val++) {
        if (len % val != 0) {
          continue;
        }
        for (int i = 0; i < val; i++) {
          int elem = cycle[i];
          int col = c[elem];
          bool ok = true;
          int ind = i;
          while (ind + val < len) {
            ind += val;
            elem = cycle[ind];
            if (col != c[elem]) {
              ok = false;
              break;
            }
          }
          if (ok) {
            _min = min(_min, val);
            break;
          }
        }
      }
    }
    cout << _min << endl;
  };
  return 0;
}
