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 T[12] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048};
int convert(string s) {
  int l = s.length(), ans = 0;
  for (int i = 0; i < l; i++) ans = 2 * ans + 1 + (s[i] - '0');
  return ans;
}
int main() {
  int n, m, l, a, b;
  cin >> n;
  vector<string> all;
  string s;
  int(*check)[2048] = new int[200][2048];
  for (int i = 0; i < 200; i++) {
    for (int j = 0; j < 2048; j++) check[i][j] = 0;
    check[i][0] = 1;
  }
  for (int i = 0; i < n; i++) {
    cin >> s;
    all.push_back(s);
  }
  for (int i = 0; i < n; i++) {
    l = all[i].length();
    for (int j = 0; j < l; j++) {
      for (int k = j; k < l; k++) {
        if (k - j + 1 <= 10) {
          a = convert(all[i].substr(j, k - j + 1));
          if (check[i][a] == 0) check[i][a]++;
        }
      }
    }
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    l = n + i;
    all.push_back(all[a - 1] + all[b - 1]);
    int c = all[a - 1].length();
    int d = all[b - 1].length();
    if (c + d > 30)
      all[l] = all[l].substr(0, 10) + all[l].substr(c + d - 10, 10);
    int x = min(9, c);
    int y = min(9, d);
    int e;
    for (int j = 1; j <= x; j++) {
      for (int k = 1; k <= y; k++) {
        if (j + k <= 10) {
          e = convert(all[a - 1].substr(c - j, j) + all[b - 1].substr(0, k));
          if (check[l][e] == 0) check[l][e]++;
        }
      }
    }
    for (int j = 0; j < 2048; j++) {
      if (check[a - 1][j] + check[b - 1][j] > 0 && check[l][j] == 0)
        check[l][j]++;
    }
    int ans = 0, cnt = 0;
    for (int j = 0; j < 2048; j++) {
      cnt = j;
      if (check[l][j] == 0) break;
    }
    for (int i = 0; i < 12; i++) {
      ans = i;
      if (T[i] > cnt + 1) break;
    }
    cout << ans - 2 << endl;
  }
  return 0;
}
