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 maxn = 105;
int n, m;
string s[maxn * 2];
set<string> var[10][maxn * 2];
void ReadInput() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> s[i];
  for (int i = 0; i < n; ++i)
    for (int j = 1; j < 10; ++j)
      for (int l = 0; l + j - 1 < (int)s[i].length(); ++l)
        var[j][i].insert(s[i].substr(l, j));
}
void Solve() {
  cin >> m;
  for (int i = n; i < n + m; ++i) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    s[i] = s[x] + s[y];
    for (int j = 1; j < 10; ++j) {
      var[j][i] = var[j][x];
      var[j][i].insert(var[j][y].begin(), var[j][y].end());
      for (int l = max((int)s[x].size() - j, 0);
           l < (int)s[x].size() && l + j - 1 < (int)s[i].size(); ++l)
        var[j][i].insert(s[i].substr(l, j));
    }
    if ((int)s[i].size() > 30)
      s[i] = s[i].substr(0, 11) + s[i].substr((int)s[i].size() - 11, 11);
    for (int j = 1; j <= 10; ++j)
      if (j == 10 || (int)var[j][i].size() != (1 << j)) {
        cout << j - 1 << '\n';
        break;
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ReadInput();
  Solve();
}