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;
const int N = 205;
string s[N];
int n, m;
int ans[N];
int fun(string x) {
  for (int i = 6; i >= 1; --i) {
    bool f = 1;
    for (int j = 0; j < (1 << i); ++j) {
      string t;
      for (int k = 0; k < i; ++k) {
        if (j & (1 << k))
          t += '1';
        else
          t += '0';
      }
      if (x.find(t) == string::npos) {
        f = 0;
        break;
      }
    }
    if (f) return i;
  }
  return 0;
}
int main() {
  memset(ans, 0, sizeof(ans));
  int i, j, a, b;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) cin >> s[i];
  scanf("%d", &m);
  for (i = n + 1; i <= n + m; ++i) {
    scanf("%d%d", &a, &b);
    s[i] = s[a] + s[b];
    if (s[i].length() > 1000) {
      string t = s[i].substr(0, 500) + s[i].substr(s[i].length() - 500, 500);
      s[i] = t;
    }
    ans[i] = max(fun(s[i]), max(ans[a], ans[b]));
    printf("%d\n", ans[i]);
  }
  return 0;
}