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;
map<int, int> mp[233][24];
map<int, int>::iterator it;
string s, sst[233], sen[233];
int n, l, t, m, p1, p2, ans;
void cpy(int now) {
  for (int i = 1; i <= 22; i++) {
    for (it = mp[p2][i].begin(); it != mp[p2][i].end(); it++) {
      mp[now][i][it->first] = 1;
    }
  }
  swap(p1, p2);
  for (int i = 1; i <= 22; i++) {
    for (it = mp[p2][i].begin(); it != mp[p2][i].end(); it++) {
      mp[now][i][it->first] = 1;
    }
  }
  swap(p1, p2);
}
int main() {
  cin >> n;
  for (int ttt = 1; ttt <= n; ttt++) {
    cin >> s;
    l = s.length();
    for (int i = 0; i < l; i++) {
      t = 0;
      for (int j = i; j < l; j++) {
        t = t * 2 + s[j] - '0';
        if (j - i + 1 > 22) break;
        mp[ttt][j - i + 1][t] = 1;
      }
    }
    if (l >= 22) {
      sst[ttt] = s.substr(0, 22);
      sen[ttt] = s.substr(l - 22, 22);
    } else {
      sst[ttt] = sen[ttt] = s;
    }
  }
  cin >> m;
  for (int ttt = n + 1; ttt <= n + m; ttt++) {
    cin >> p1 >> p2;
    cpy(ttt);
    s = sen[p1] + sst[p2];
    l = s.length();
    for (int i = 0; i < l; i++) {
      t = 0;
      for (int j = i; j < l; j++) {
        t = t * 2 + s[j] - '0';
        if (j - i + 1 > 22) break;
        mp[ttt][j - i + 1][t] = 1;
      }
    }
    ans = 0;
    for (int i = 1; i <= 22; i++) {
      if (mp[ttt][i].size() == (1 << i)) {
        ans = max(ans, i);
      }
    }
    printf("%d\n", ans);
    s = sst[p1] + sst[p2];
    sst[ttt] = s.substr(0, min(22, (int)s.length()));
    s = sen[p1] + sen[p2];
    sen[ttt] = s.substr(max((int)s.length() - 22, 0), min(22, (int)s.length()));
  }
}