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;
string a[210];
bool h[20000];
int res[20000];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    l--, r--;
    a[n + i] = a[l] + a[r];
    if (a[n + i].length() > 2000)
      a[n + i] = a[n + i].substr(0, 1000) +
                 a[n + i].substr(a[n + i].length() - 1000, 1000);
    int len = a[n + i].size();
    int ans = 0;
    for (int j = 1; j < 15; j++) {
      memset(h, 0, sizeof(h));
      int temp = 0;
      int ss = 1 << (j - 1);
      for (int k = 0; k < len; k++) {
        if (k + 1 <= j) {
          temp <<= 1;
          temp += a[n + i][k] - '0';
        } else {
          h[temp] = true;
          temp ^= (ss * (a[n + i][k - j] - '0'));
          temp <<= 1;
          temp += a[n + i][k] - '0';
        }
      }
      h[temp] = true;
      bool flag = true;
      for (int k = 0; k < ss * 2; k++) {
        if (!h[k]) {
          flag = false;
          break;
        }
      }
      if (flag)
        ans = j;
      else
        break;
    }
    res[n + i] = max(max(res[l], res[r]), ans);
    printf("%d\n", res[n + i]);
  }
  return 0;
}