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;
using LL = long long;
constexpr LL LINF = 334ll << 53;
constexpr int INF = 15 << 26;
constexpr LL MOD = 1E9 + 7;
struct Problem {
  int n, q;
  vector<string> s, tail;
  vector<vector<vector<int>>> dp;
  Problem(LL n)
      : n(n), s(n), dp(200, vector<vector<int>>(10, vector<int>((1 << 9)))){};
  void solve() {
    for (int i = 0; i < n; ++i) {
      cin >> s[i];
      tail.push_back(s[i]);
      for (int j = 1; j <= min((int)s[i].size(), 9); ++j) {
        int tmp = 0;
        for (int k = 0; k < j; ++k) {
          tmp <<= 1;
          tmp += s[i][k] - '0';
        }
        dp[i][j][tmp] = 1;
        for (int k = j; k < (int)s[i].size(); ++k) {
          tmp <<= 1;
          if ((tmp >> j) & 1) tmp ^= (1 << j);
          tmp += s[i][k] - '0';
          dp[i][j][tmp] = 1;
        }
      }
    }
    cin >> q;
    for (int i = 0; i < q; ++i) {
      int a, b;
      cin >> a >> b;
      --a;
      --b;
      for (int j = 1; j <= 9; ++j) {
        for (int k = 0; k < (1 << j); ++k) {
          dp[n + i][j][k] = dp[a][j][k] | dp[b][j][k];
        }
      }
      s.push_back(s[a] + s[b]);
      s[i + n] = s[i + n].substr(0, 10);
      tail.push_back(tail[a] + tail[b]);
      tail[i + n] = tail[i + n].substr(max(0, (int)tail[i + n].size() - 10));
      for (int j = 2; j <= min(9, (int)s[a].size() + (int)s[b].size()); ++j) {
        string ns = tail[a].substr(max(0, (int)tail[a].size() - j + 1));
        int len = (int)ns.size();
        ns += s[b].substr(0, j - len);
        int tmp = 0;
        for (int k = 0; k < j; ++k) {
          tmp <<= 1;
          tmp += ns[k] - '0';
        }
        dp[n + i][j][tmp] = 1;
        for (int k = 1; k < len; ++k) {
          ;
          tmp <<= 1;
          if ((tmp >> j) & 1) tmp ^= (1 << j);
          if (j - len + k - 1 >= (int)s[b].size()) break;
          tmp += s[b][j - len + k - 1] - '0';
          dp[n + i][j][tmp] = 1;
        }
      }
      int ans = 0;
      for (int j = 1; j <= 9; ++j) {
        if (accumulate(dp[n + i][j].begin(), dp[n + i][j].end(), 0) < (1 << j))
          break;
        ans = j;
      }
      cout << ans << "\n";
    }
  }
};
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long n = 0;
  cin >> n;
  Problem p(n);
  p.solve();
  return 0;
}