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;
int64_t mod = 1e9 + 7, inf = 4e18;
int64_t bruteForce(string &s) {
  for (int64_t i = 1; i <= 10; i++) {
    for (int64_t j = 0; j < (1 << i); j++) {
      string cur = "";
      for (int64_t k = 0; k < i; k++) cur += (j & (1 << k)) ? "1" : "0";
      if (s.find(cur) == -1) return i - 1;
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  cout << fixed << setprecision(0);
  int64_t n, m;
  cin >> n;
  vector<int64_t> ans(n);
  vector<string> s(n);
  for (int64_t i = 0; i < int64_t(n); ++i) {
    cin >> s[i];
    ans[i] = bruteForce(s[i]);
  }
  int64_t l, r, ind = 0;
  for (cin >> m; m--; cout << ans.back() << endl) {
    cin >> l >> r;
    l--, r--;
    s.push_back(s[l] + s[r]);
    if (s.back().size() > 1000) {
      s.back() =
          s.back().substr(0, 500) + s.back().substr(s.back().size() - 500, 500);
    }
    ans.push_back(max(bruteForce(s.back()), max(ans[l], ans[r])));
  }
  return 0;
}