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 N = 205, M = 2000;
int n, m, x, y, ans[N];
string s[N];
inline bool check(string &s, int k) {
  int n = s.size();
  set<string> q;
  for (int i = 0; i + k - 1 < n; i++) {
    q.insert(s.substr(i, k));
  }
  return q.size() == (1 << k);
}
inline int get(string &s) {
  int k = 1;
  while (check(s, k)) {
    k++;
  }
  return k - 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    ans[i] = get(s[i]);
  }
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    cin >> x >> y;
    s[i] = s[x] + s[y];
    int besit = s[i].size();
    if (besit > M) {
      s[i] = s[i].substr(0, M / 2) + s[i].substr(besit - M / 2);
    }
    ans[i] = max({get(s[i]), ans[x], ans[y]});
    cout << ans[i] << "\n";
  }
}