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;
int n, m, tt[222], tini[222], tfim[222];
string ini[222], fim[222];
char s[222][222];
set<string> tem[222][22];
int32_t main() {
  scanf("%d", &n);
  for (int(i) = (0); (i) < (n); (i)++) {
    scanf("%s", s[i]);
    ini[i] = fim[i] = string(s[i]);
    tini[i] = tfim[i] = tt[i] = strlen(s[i]);
    for (int(j) = (0); (j) < (tt[i]); (j)++) {
      string cur = "";
      for (int(k) = (j); (k) < (tt[i]); (k)++) {
        cur += s[i][k];
        if (((int)(cur).size()) >= 22) break;
        tem[i][((int)(cur).size())].insert(cur);
      }
    }
  }
  scanf("%d", &m);
  for (int(qq) = (0); (qq) < (m); (qq)++) {
    int a, b, c = qq + n;
    scanf("%d%d", &a, &b);
    a--, b--;
    ini[c] = ini[a];
    int ptr = 0;
    while (((int)(ini[c]).size()) < 22 && ptr < ((int)(ini[b]).size()))
      ini[c] += ini[b][ptr], ptr++;
    ptr = 0;
    fim[c] = fim[b];
    reverse((fim[c]).begin(), (fim[c]).end());
    while (((int)(fim[c]).size()) < 22 && ptr < ((int)(fim[a]).size()))
      fim[c] += fim[a][((int)(fim[a]).size()) - 1 - ptr], ptr++;
    reverse((fim[c]).begin(), (fim[c]).end());
    for (int(k) = (0); (k) < (22); (k)++) {
      if (((int)(tem[a][k]).size()) > ((int)(tem[b][k]).size())) {
        tem[c][k] = tem[a][k];
        for (string ss : tem[b][k]) {
          tem[c][k].insert(ss);
          if (((int)(tem[c][k]).size()) == (1 << k)) break;
        }
      } else {
        tem[c][k] = tem[b][k];
        for (string ss : tem[a][k]) {
          tem[c][k].insert(ss);
          if (((int)(tem[c][k]).size()) == (1 << k)) break;
        }
      }
    }
    for (int(k) = (0); (k) < (22); (k)++)
      for (int(i) = (0); (i) < (((int)(fim[a]).size())); (i)++) {
        int tl = ((int)(fim[a]).size()) - i, tr = k - tl;
        if (tr <= 0 || tr > ((int)(ini[b]).size())) continue;
        string cur = "";
        for (int(j) = (i); (j) < (((int)(fim[a]).size())); (j)++)
          cur += fim[a][j];
        for (int(j) = (0); (j) < (tr); (j)++) cur += ini[b][j];
        tem[c][k].insert(cur);
      }
    int ans = 0;
    for (int(i) = (1); (i) < (22); (i)++)
      if (((int)(tem[c][i]).size()) == (1 << i)) ans = i;
    printf("%d\n", ans);
  }
}