Problem: You are given n strings s1, s2, ..., sn consisting of characters 0 and 1. m operations are performed, on each of them you concatenate two existing strings into a new one. On the i-th operation the concatenation saisbi is saved into a new string sn + i (the operations are numbered starting from 1). After each operation you need to find the maximum positive integer k such that all possible strings consisting of 0 and 1 of length k (there are 2k such strings) are substrings of the new string. If there is no such k, print 0.

Input

The first line contains single integer n (1 ≤ n ≤ 100) — the number of strings. The next n lines contain strings s1, s2, ..., sn (1 ≤ |si| ≤ 100), one per line. The total length of strings is not greater than 100.

The next line contains single integer m (1 ≤ m ≤ 100) — the number of operations. m lines follow, each of them contains two integers ai abd bi (1 ≤ ai, bi ≤ n + i - 1) — the number of strings that are concatenated to form sn + i.

Output

Print m lines, each should contain one integer — the answer to the question after the corresponding operation.

Example

Input

5
01
10
101
11111
0
3
1 2
6 5
4 4


Output

1
2
0

Note

On the first operation, a new string "0110" is created. For k = 1 the two possible binary strings of length k are "0" and "1", they are substrings of the new string. For k = 2 and greater there exist strings of length k that do not appear in this string (for k = 2 such string is "00"). So the answer is 1.

On the second operation the string "01100" is created. Now all strings of length k = 2 are present.

On the third operation the string "1111111111" is created. There is no zero, so the answer is 0.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int k(string s) {
  for (int i = 9; i; i--) {
    bool y = 0;
    for (int j = 0; j < (1 << i); j++) {
      string t;
      for (int l = 0; l < i; l++) {
        if (j & (1 << l))
          t += '1';
        else
          t += '0';
      }
      if (s.find(t) == string::npos) {
        y = 1;
        break;
      }
    }
    if (!y) return i;
  }
  return 0;
}
int main() {
  int n, m, a, b, z[202] = {};
  string s[202] = {};
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> s[i];
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    cin >> a >> b;
    s[i] = s[a] + s[b];
    int L = 510;
    if (s[i].length() > L + L) {
      string t = s[i].substr(0, L) + s[i].substr(s[i].length() - L, L);
      s[i] = t;
    }
    z[i] = max({z[a], z[b], k(s[i])});
    cout << z[i] << '\n';
  }
}