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 km = 10;
int main() {
  int n;
  cin >> n;
  vector<string> pref(n), post(n);
  vector<vector<unordered_set<string>>> subs(
      n, vector<unordered_set<string>>(km + 1));
  for (int i = 0; i < n; ++i) {
    string str;
    cin >> str;
    pref[i] = str.substr(0, km);
    post[i] = str.substr(max(0, (int)(str).size() - km), km);
    for (int k = 1; k <= km; ++k)
      for (int j = 0; j < (int)(str).size() - k + 1; ++j)
        subs[i][k].insert(str.substr(j, k));
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int ai, bi;
    cin >> ai >> bi;
    ai--;
    bi--;
    pref.push_back((pref[ai] + pref[bi]).substr(0, km));
    post.push_back(
        (post[ai] + post[bi])
            .substr(max(0, (int)(post[ai] + post[bi]).size() - km), km));
    subs.push_back(vector<unordered_set<string>>(km + 1));
    string str = post[ai] + pref[bi];
    int res = 0;
    for (int k = 1; k <= km; ++k) {
      subs.back()[k].insert(subs[ai][k].begin(), subs[ai][k].end());
      subs.back()[k].insert(subs[bi][k].begin(), subs[bi][k].end());
      for (int j = 0; j < (int)(str).size() - k + 1; ++j)
        subs.back()[k].insert(str.substr(j, k));
      if ((int)(subs.back()[k]).size() == (1 << k)) res = k;
    }
    cout << res << endl;
  }
}