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;
const int N = int(6e5) + 5;
const int inf = (int)1e9 + 7;
const int maxk = 13;
int n;
int wow[(1 << maxk)], ans[201];
string s[201];
int find_num(string x) {
  for (int l = 1; l <= 9; ++l) {
    for (int j = 0; j < (1 << l); ++j) {
      string now;
      for (int k = 0; k < l; ++k) {
        now += (j >> k) & 1 ? '1' : '0';
      }
      if (x.find(now) == string ::npos) {
        return l - 1;
      }
    }
  }
}
void connect(int a, int b, int t) {
  ans[t] = max(ans[a], ans[b]);
  s[t] = s[a] + s[b];
  if ((int)s[t].size() > 1000) {
    s[t] = s[t].substr(0, 500) + s[t].substr((int)s[t].size() - 500, 500);
  }
  ans[t] = max(ans[t], find_num(s[t]));
}
int main() {
  cin >> n;
  int sz = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> s[i];
    ans[i] = find_num(s[i]);
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    int a, b;
    cin >> a >> b;
    connect(a, b, i + n);
    cout << ans[i + n] << "\n";
  }
  return 0;
}