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;
string s[205][2];
int ans[205];
int n, m;
int calc(string x) {
  int val = 1;
  while (true) {
    bool ok = 1;
    for (int k = 0; k < (1 << val); k++) {
      string go;
      for (int i = 0; i < val; i++) go.push_back('0' + bool(k & (1 << i)));
      if (x.find(go) == string::npos) {
        ok = 0;
        break;
      }
    }
    if (!ok) return val - 1;
    val++;
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i][0];
    s[i][1] = s[i][0];
    ans[i] = calc(s[i][0]);
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    s[n + i][0] = s[x][0] + s[y][0];
    s[n + i][1] = s[x][1] + s[y][1];
    reverse(s[n + i][1].begin(), s[n + i][1].end());
    s[n + i][0] = s[n + i][0].substr(
        0, (int(s[n + i][0].size()) < 1000 ? int(s[n + i][0].size()) : 1000));
    s[n + i][1] = s[n + i][1].substr(
        0, (int(s[n + i][1].size()) < 1000 ? int(s[n + i][1].size()) : 1000));
    reverse(s[n + i][1].begin(), s[n + i][1].end());
    ans[n + i] = max(ans[x], ans[y]);
    ans[n + i] = max(ans[n + i], calc(s[x][1] + s[y][0]));
    cout << ans[n + i] << '\n';
  }
  return 0;
}