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 INF = 1e9 + 10, MN = 222, MS = 1055;
string s[MN], L[MN], R[MN];
int sz[MN], n, m;
bitset<MS> Bt[MN][11];
void append(int id, const string &s) {
  int m = ((int)(s).size());
  for (int i = 0; i < m; ++i) {
    int cur = 0;
    for (int j = i; j < i + 11 && j < m; ++j) {
      cur = cur * 2 + (s[j] == '1');
      Bt[id][j - i][cur] = 1;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> s[i], sz[i] = ((int)(s[i]).size());
    L[i] = R[i] = s[i];
    append(i, s[i]);
  }
  cin >> m;
  for (int i = n; i < n + m; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    sz[i] = sz[x] + sz[y];
    for (int j = 0; j < 11; ++j) Bt[i][j] = Bt[x][j] | Bt[y][j];
    append(i, R[x].substr(max(((int)(R[x]).size()) - 10, 0),
                          min(10, ((int)(R[x]).size()))) +
                  L[y].substr(0, min(10, ((int)(L[y]).size()))));
    {
      L[i] = L[x];
      int cnt = 0;
      while (((int)(L[i]).size()) <= 11 && cnt < ((int)(L[y]).size()))
        L[i] += L[y][cnt++];
    }
    {
      R[i] = R[y];
      int cnt = ((int)(R[x]).size()) - 1;
      while (((int)(R[i]).size()) <= 11 && ~cnt) R[i] = R[x][cnt--] + R[i];
    }
    if (sz[i] > INF) sz[i] = INF;
    for (int k = 10;; --k) {
      if (k == 0) {
        cout << 0 << '\n';
        break;
      }
      bool Fail = false;
      for (int msk = 0; msk < (1 << k); ++msk) {
        if (!Bt[i][k - 1][msk]) Fail = true;
      }
      if (!Fail) {
        cout << k << '\n';
        break;
      }
    }
  }
  return 0;
}