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 MAXN = 300 + 10;
string str[MAXN];
int ans[MAXN];
int solve(int n) {
  for (int i = 1; i <= 10; ++i) {
    for (int k = 0; k <= (1 << i); ++k) {
      string s = "";
      for (int j = i - 1; j >= 0; --j) {
        if ((1 << j) & k)
          s += '1';
        else
          s += '0';
      }
      if (str[n].find(s) == -1) {
        return i - 1;
      }
    }
  }
}
int main(void) {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> str[i];
  }
  memset(ans, 0x00, sizeof(ans));
  int m;
  cin >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    str[++n] = str[a] + str[b];
    if (str[n].size() > 1000) {
      str[n] = str[n].substr(0, 500) + str[n].substr(str[n].size() - 500, 500);
    }
    ans[n] = max(solve(n), max(ans[a], ans[b]));
    cout << ans[n] << endl;
  }
}