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;
vector<string> str(1);
vector<int> ans(1);
int solve(const string& s) {
  for (int len = 1; len <= 10; ++len) {
    for (int opt = 0; opt < (1 << len); ++opt) {
      bitset<10> bits(opt);
      string tmpS(len, '0');
      for (int i = 0; i < len; ++i) {
        tmpS[i] += bits[i];
      }
      if (s.find(tmpS) == string::npos) {
        return len - 1;
      }
    }
  }
  return -1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    char buff[256];
    scanf("%s", buff);
    str.push_back(buff);
    ans.push_back(solve(str.back()));
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    str.push_back(str[a] + str[b]);
    if (str.back().length() > 1000) {
      str.back() = str.back().substr(0, 500) +
                   str.back().substr(str.back().length() - 500, 500);
    }
    ans.push_back(max({solve(str.back()), ans[a], ans[b]}));
    cout << ans.back() << endl;
  }
  return 0;
}
