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, cnt, ans, fo[210][20], la[210][20], len[210];
int bi[30], tot;
char ss[210];
bool flag, mp[210][20010];
void upd(int a, int b, int c) {
  int i, s, j, k;
  for (i = 1; i <= min(len[b], 12); i++) fo[a][i] = fo[b][i];
  s = 1;
  k = len[b];
  while (k < 12 && s <= len[c]) fo[a][++k] = fo[c][s++];
  for (i = 1; i <= min(len[c], 12); i++) la[a][i] = la[c][i];
  s = 1;
  k = len[c];
  while (k < 12 && s <= len[b]) la[a][++k] = la[b][s++];
  len[a] = min(len[b] + len[c], 12);
  for (i = 0; i < bi[14]; i++) mp[a][i] = (mp[b][i] | mp[c][i]);
  tot = 0;
  for (s = 1; s <= len[b]; s++) {
    tot += la[b][s] * bi[s - 1];
    j = 0;
    for (k = 1; k <= len[c] && s + k <= 12; k++)
      j = (j << 1) + fo[c][k], mp[a][tot * bi[k] + j + bi[s + k]] = 1;
  }
}
int main() {
  int i, s, a, b, j, k;
  bi[0] = 1;
  for (i = 1; i <= 20; i++) bi[i] = bi[i - 1] << 1;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf(" %s", ss);
    len[i] = min((int)strlen(ss), 12);
    k = strlen(ss);
    for (s = 1; s <= len[i]; s++)
      fo[i][s] = ss[s - 1] - '0', la[i][s] = ss[k - s] - '0';
    for (s = strlen(ss) - 1; s >= 0; s--) {
      tot = 0;
      for (k = s; k >= max(0, s - 12); k--) {
        j = ss[k] - '0';
        tot += bi[s - k] * j;
        mp[i][tot + bi[s - k + 1]] = 1;
      }
    }
  }
  scanf("%d", &m);
  for (i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    upd(++n, a, b);
    ans = 0;
    for (s = 12; s && !ans; s--) {
      ans = s;
      for (k = 0; k < bi[s] && ans; k++)
        if (!mp[n][k + bi[s]]) ans = 0;
    }
    printf("%d\n", ans);
  }
  return 0;
}