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 int N = 404;
string c[N], s;
int ans[N], n, m;
int Brute_Force() {
  for (int i = 1; i <= 10; i++)
    for (int k = 0; k <= 1 << i; k++) {
      s = "";
      for (int j = 0; j <= i - 1; j++) (1 << j)& k ? s += '1' : s += '0';
      if (c[n].find(s) == -1) return i - 1;
    }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> c[i], ans[i] = 0;
  cin >> m;
  while (m--) {
    int a, b;
    cin >> a >> b;
    c[++n] = c[a] + c[b];
    if (c[n].size() > 1000)
      c[n] = c[n].substr(0, 500) + c[n].substr(c[n].size() - 500, 500);
    ans[n] = max(Brute_Force(), max(ans[a], ans[b]));
    cout << ans[n] << endl;
  }
  return 0;
}
