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;
int n, m;
string str[205];
bool flag[205][15][1 << 14];
int main(void) {
  scanf("%d", &n);
  memset(flag, false, sizeof(flag));
  for (int i = 0; i < n; i++) {
    cin >> str[i];
    int len = str[i].size();
    for (int j = 0; j < len; j++) {
      int bit = 0;
      for (int k = 0; k < 14; k++) {
        if (j + k < len) {
          bit *= 2;
          if (str[i][j + k] == '1') bit++;
          flag[i][k + 1][bit] = true;
        }
      }
    }
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    int len = str[a].size();
    int len2 = str[b].size();
    str[n + i] = str[a] + str[b];
    if (str[n + i].size() > 500) {
      if (len < 50) {
        str[n + i] = str[a] + str[b].substr(0, 50) + str[b].substr(len2 - 30);
      } else if (len2 < 50) {
        str[n + i] = str[a].substr(0, 50) + str[a].substr(len - 30) + str[b];
      } else {
        str[n + i] = str[a].substr(0, 50) + str[b].substr(len2 - 50);
      }
    }
    for (int k = 1; k < 15; k++) {
      for (int l = 0; l < (1 << k); l++) {
        if (flag[a][k][l]) flag[n + i][k][l] = true;
        if (flag[b][k][l]) flag[n + i][k][l] = true;
      }
      for (int j = 1; j <= min(k - 1, len); j++) {
        int bit = 0;
        int l2 = k - j;
        if (l2 > len2) continue;
        for (int lb = 0; lb < j; lb++) {
          if (str[a][len - 1 - lb] == '1') bit += (1 << lb);
        }
        for (int lb = 0; lb < l2; lb++) {
          bit *= 2;
          if (str[b][lb] == '1') bit++;
        }
        flag[n + i][k][bit] = true;
      }
    }
    int ans = 0;
    for (int j = 1; j <= 14; j++) {
      bool fl = true;
      for (int k = 0; k < (1 << j); k++) {
        if (!flag[n + i][j][k]) {
          fl = false;
          break;
        }
      }
      if (!fl) {
        break;
      } else {
        ans++;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}