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 dp[205][14][20000];
char a[205][20000];
int ans[20000];
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", a[i]);
    int len = strlen(a[i]);
    char *tmp = a[i];
    for (int j = 0; j < len; ++j) {
      int now = 0;
      for (int k = j; k < len; ++k) {
        if (k - j + 1 == 14) break;
        now = (now << 1) + tmp[k] - '0';
        dp[i][k - j + 1][now]++;
      }
    }
  }
  int m;
  scanf("%d", &m);
  int fir, sec;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &fir, &sec);
    for (int p = 1; p <= 14; ++p) {
      for (int j = 0; j <= (1 << p); ++j) {
        if (dp[fir][p][j] || dp[sec][p][j]) dp[n + i][p][j] = 1;
      }
    }
    char tmp[305];
    int lena = strlen(a[fir]);
    int lenb = strlen(a[sec]);
    int q = 0;
    for (int j = max(0, lena - 14); j < lena; ++j) {
      tmp[q++] = a[fir][j];
    }
    for (int j = 0; j < min(lenb, 14); ++j) {
      tmp[q++] = a[sec][j];
    }
    tmp[q] = '\0';
    for (int j = 0; j < q; ++j) {
      int now = 0;
      for (int k = j; k < q; ++k) {
        if (k - j + 1 == 14) break;
        now = (now << 1) + tmp[k] - '0';
        dp[n + i][k - j + 1][now]++;
      }
    }
    int flag = 0;
    for (int p = 1; p <= 14; ++p) {
      for (int j = 0; j <= (1 << p) - 1; ++j) {
        if (dp[n + i][p][j] == 0) {
          flag = 1;
          printf("%d\n", p - 1);
          break;
        }
      }
      if (flag) break;
    }
    q = 0;
    if (lena + lenb < 14) {
      for (int j = 0; j < lena; ++j) {
        a[n + i][q++] = a[fir][j];
      }
      for (int j = 0; j < lenb; ++j) {
        a[n + i][q++] = a[sec][j];
      }
    } else {
      for (int j = 0; j < min(14, lena); ++j) {
        a[n + i][q++] = a[fir][j];
      }
      if (lena < 14)
        for (int j = 0; j < max(0, (14 - lena)); ++j) {
          a[n + i][q++] = a[sec][j];
        }
      for (int j = max(0, lena - (14 - lenb)); j < lena; ++j) {
        a[n + i][q++] = a[fir][j];
      }
      for (int j = max(0, lenb - 14); j < lenb; ++j) {
        a[n + i][q++] = a[sec][j];
      }
    }
    a[n + i][q] = '\0';
  }
}