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>
int n, q;
char s[110][110];
int dp[210][12][4096];
char cs[210][15];
char ce[210][15];
int clen[210];
char stmp[40];
void calcdp(int x, char *row) {
  int m = strlen(row);
  for (int i = 0; i < 12; i++) {
    for (int j = 0; j < (1 << (i + 1)); j++) {
      for (int k = 0; k + i < m; k++) {
        int sflg = 0;
        for (int ii = 0; ii <= i; ii++) {
          if (((j >> ii) & 1) != row[k + ii] - '0') {
            sflg = 1;
            break;
          }
        }
        if (sflg == 0) {
          dp[x][i][j] = 1;
          break;
        }
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s[i]);
    calcdp(i, s[i]);
    int sl = strlen(s[i]);
    clen[i] = sl;
    if (clen[i] > 14) clen[i] = 14;
    for (int j = 0; j < clen[i]; j++) {
      cs[i][j] = s[i][j];
      ce[i][j] = s[i][sl - clen[i] + j];
    }
  }
  scanf("%d", &q);
  for (int i = n; i < n + q; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    for (int j = 0; j < 12; j++) {
      for (int k = 0; k < (1 << (j + 1)); k++) {
        dp[i][j][k] = dp[a][j][k] | dp[b][j][k];
      }
    }
    for (int j = 0; j < clen[a]; j++) {
      stmp[j] = ce[a][j];
    }
    for (int j = 0; j < clen[b]; j++) {
      stmp[j + clen[a]] = cs[b][j];
    }
    stmp[clen[a] + clen[b]] = 0;
    calcdp(i, stmp);
    clen[i] = clen[a] + clen[b];
    if (clen[i] > 14) clen[i] = 14;
    for (int j = 0; j < clen[a]; j++) {
      cs[i][j] = cs[a][j];
    }
    for (int j = 0; j + clen[a] < clen[i]; j++) {
      cs[i][j + clen[a]] = cs[b][j];
    }
    for (int j = 0; j < clen[b]; j++) {
      ce[i][j + clen[i] - clen[b]] = ce[b][j];
    }
    for (int j = 0; j + clen[b] < clen[i]; j++) {
      ce[i][j] = ce[a][j + clen[a] + clen[b] - clen[i]];
    }
    int res = 12;
    for (int j = 0; j < 12; j++) {
      int flg = 0;
      for (int k = 0; k < (1 << (j + 1)); k++) {
        if (dp[i][j][k] == 0) {
          flg = 1;
          break;
        }
      }
      if (flg == 1) {
        res = j;
        break;
      }
    }
    printf("%d\n", res);
  }
  return 0;
}