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.
#include <bits/stdc++.h>
using namespace std;
int sumk[205];
int main() {
  int n;
  cin >> n;
  vector<string> s;
  for (int i = 0; i < n; i++) {
    string x;
    cin >> x;
    s.push_back(x);
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    string news = s[a - 1] + s[b - 1];
    if (news.length() > 1000)
      news = news.substr(0, 500) + news.substr(news.length() - 500, 500);
    s.push_back(news);
    int k = 1;
    bool flag = false;
    while (!flag && k <= 10) {
      for (int j = 0; j < (1 << k); ++j) {
        string t;
        for (int h = 0; h < k; ++h) {
          if (j & (1 << h))
            t += '1';
          else
            t += '0';
        }
        if (news.find(t) == -1) {
          flag = true;
          break;
        }
      }
      if (flag) break;
      k++;
    }
    sumk[n + i] = max(max(sumk[a - 1], sumk[b - 1]), k - 1);
    cout << sumk[n + i] << endl;
  }
  return 0;
}
