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.
#include <bits/stdc++.h>
using namespace std;
string s[1001];
pair<string, string> com[1001];
set<int> has[300][30];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; i <= int(n); i++) {
    cin >> s[i];
    int len = int(s[i].size());
    for (int k = 0; k < len; k++) {
      int val = 0;
      int l = 0;
      for (int j = k; j < len; j++) {
        l++;
        val = val * 2 + int(s[i][j] - '0');
        if (val >= 1 << 16) break;
        { has[i][l].insert(val); }
      }
    }
    if (len <= 16) {
      com[i].first = s[i];
      com[i].second = s[i];
    } else {
      com[i].first = s[i].substr(0, 16);
      com[i].second = s[i].substr(len - 16, 16);
    }
  }
  int m;
  cin >> m;
  for (int k = 1; k <= int(m); k++) {
    int x, y;
    cin >> x >> y;
    string a = com[x].second;
    string b = com[y].first;
    string res = a + b;
    int len1 = int(com[x].second.size());
    int len2 = int(com[y].first.size());
    for (int i = 1; i <= len1; i++) {
      for (int j = 1; j <= len2; j++) {
        int total = i + j;
        string fir = com[x].second.substr(len1 - i);
        string sec = com[y].first.substr(0, j);
        string result = fir + sec;
        int val = 0;
        if (total > 16) continue;
        for (int i = 0; i < total; i++) {
          val = val * 2 + int(result[i] - '0');
        }
        has[k + n][total].insert(val);
      }
    }
    for (int i = 1; i <= 16; i++) {
      for (auto aa : has[x][i]) has[k + n][i].insert(aa);
      for (auto aa : has[y][i]) has[k + n][i].insert(aa);
    }
    int ans = 0;
    for (int i = 1; i <= 16; i++) {
      if (int(has[k + n][i].size()) == (1 << i))
        ans = i;
      else
        break;
    }
    cout << ans << '\n';
    if (int(com[x].first.size()) >= 16) {
      com[k + n].first = com[x].first;
    } else {
      assert(com[x].first == com[x].second);
      int rem = 16 - int(com[x].first.size());
      string more = com[y].first.substr(0, min(int(com[y].first.size()), rem));
      com[k + n].first = com[x].first + more;
    }
    if (int(com[y].second.size()) >= 16) {
      com[k + n].second = com[y].second;
    } else {
      assert(com[y].first == com[y].second);
      int rem = 16 - int(com[y].second.size());
      int len1 = int(com[x].second.size());
      int cando = min(rem, len1);
      string more = com[x].second.substr(len1 - cando, cando);
      com[k + n].second = more + com[y].second;
    }
  }
}
