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 MN = 200;
vector<string> str(MN + 5);
vector<set<string> > sub(MN + 5);
int N, K;
void get_sub(const string& s, set<string>& sub) {
  for (int i = 0; i < s.size(); i++) {
    for (int j = 1; j <= min((int)s.size(), 10); j++) {
      if (i + j > s.size()) break;
      sub.insert(s.substr(i, j));
    }
  }
}
void merge_sub(set<string>& to, set<string>& from) {
  for (auto& v : from) {
    to.insert(v);
  }
}
int solve(set<string>& sub) {
  for (int i = 9; i >= 1; i--) {
    bool flag = 1;
    for (int j = 0; j < (1 << i); j++) {
      string tmp = "";
      for (int k = 0; k < i; k++) {
        if (j & (1 << k))
          tmp += '1';
        else
          tmp += '0';
      }
      if (sub.find(tmp) == sub.end()) {
        flag = 0;
        break;
      }
    }
    if (flag) return i;
  }
  return 0;
}
int main() {
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> str[i];
    get_sub(str[i], sub[i]);
  }
  cin >> K;
  for (int i = N + 1; i <= N + K; i++) {
    int a, b;
    cin >> a >> b;
    merge_sub(sub[i], sub[a]);
    merge_sub(sub[i], sub[b]);
    int sz = (int)str[a].size();
    int len = min(sz, 20);
    string new_str = str[a].substr(sz - len, len);
    sz = (int)str[b].size();
    len = min(sz, 20);
    new_str += str[b].substr(0, len);
    get_sub(new_str, sub[i]);
    cout << solve(sub[i]) << endl;
    str[i] = str[a] + str[b];
    sz = (int)str[i].size();
    if (sz > 40) {
      str[i] = str[i].substr(0, 20) + str[i].substr(sz - 20, 20);
    }
  }
  return 0;
}