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 lim = 13;
int calc(const string &s, set<pair<int, int> > &vs) {
  int ans = 0;
  for (int k = lim; k >= 1; k--) {
    int found = true;
    for (int i = 0; i < (1 << k); i++) {
      if (vs.count({k, i})) continue;
      string f;
      for (int j = 0; j < k; j++) f += "01"[(i >> j) & 1];
      if (s.find(f) == string::npos)
        found = false;
      else
        vs.emplace(k, i);
    }
    if (!found) ans = k - 1;
  }
  return (ans);
}
int main() {
  int N, M;
  string S[200];
  int pre[200] = {};
  set<pair<int, int> > found[200];
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> S[i];
  }
  cin >> M;
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    S[i + N] = S[--a] + S[--b];
    for (auto &p : found[a]) found[i + N].emplace(p);
    for (auto &p : found[b]) found[i + N].emplace(p);
    pre[i + N] = max({pre[a], pre[b], calc(S[i + N], found[i + N])});
    cout << pre[i + N] << endl;
    if (S[i + N].size() > 24) {
      S[i + N] =
          S[i + N].substr(0, 12) + "#" + S[i + N].substr(S[i + N].size() - 12);
    }
  }
}