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 MAX = (1 << 12);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<pair<string, string>> hugs(n);
  vector<vector<bitset<MAX>>> codes(n, vector<bitset<MAX>>(12));
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    int sz = s.size();
    hugs[i].first = s.substr(0, min(sz, 12));
    hugs[i].second = s.substr(sz - min(sz, 12), min(sz, 12));
    for (int j = 1; j <= 12; j++) {
      for (int k = 0; k + j - 1 < sz; k++) {
        string cur = s.substr(k, j);
        int val = 0;
        for (int x = 0; x < (int)cur.size(); x++) {
          val *= 2;
          val += (cur[x] - '0');
        }
        codes[i][j - 1][val] = 1;
      }
    }
  }
  int m;
  cin >> m;
  hugs.reserve(n + m);
  codes.reserve(n + m);
  while (m--) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    vector<bitset<MAX>> nw(12);
    for (int i = 0; i < 12; i++) nw[i] = codes[l][i] | codes[r][i];
    string merge = hugs[l].second + hugs[r].first;
    int msz = merge.size();
    for (int l = 1; l <= 12; l++) {
      for (int i = 0; i + l - 1 < msz; i++) {
        string cur = merge.substr(i, l);
        int val = 0;
        for (int x = 0; x < (int)cur.size(); x++) {
          val *= 2;
          val += (cur[x] - '0');
        }
        nw[l - 1][val] = 1;
      }
    }
    string left = hugs[l].first + hugs[r].first;
    string right = hugs[l].second + hugs[r].second;
    int lsz = left.size(), rsz = right.size();
    pair<string, string> Ts = {left.substr(0, min(lsz, 12)),
                               right.substr(rsz - min(rsz, 12), min(rsz, 12))};
    hugs.emplace_back(Ts);
    codes.emplace_back(nw);
    int mx = 0;
    for (int i = 0; i < 12; i++) {
      int lim = 1 << (i + 1);
      int ok = 1;
      for (int j = 0; j < lim; j++) {
        if (!nw[i][j]) ok = 0;
      }
      if (ok) mx = max(mx, i + 1);
    }
    cout << mx << '\n';
  }
  return 0;
}