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.
#include <bits/stdc++.h>
using namespace std;
const long long N = 205;
string arr[N];
long long ans[N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  long long m;
  cin >> m;
  auto get = [&](long long a) {
    for (long long i = 1; i <= 10; i++) {
      for (long long j = 0; j < (1 << i); j++) {
        string s(i, '0');
        for (long long k = 0; k < i; k++) {
          if (j & (1 << k)) s[k] = '1';
        }
        if (arr[a].find(s) == -1) return i - 1;
      }
    }
  };
  for (long long i = 1; i <= m; i++) {
    long long l, r;
    cin >> l >> r;
    arr[i + n] = arr[l] + arr[r];
    ans[i + n] = max({ans[l], ans[r], get(i + n)});
    if (arr[i + n].size() > 2000)
      arr[i + n] = arr[i + n].substr(0, 1000) +
                   arr[i + n].substr(arr[i + n].size() - 1000);
    cout << ans[i + n] << "\n";
  }
}
