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;
int n, m, x, y, i, j;
set<int> S[222][222];
string L[222], R[222], s[222];
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    cin >> s[i];
    L[i] = R[i] = s[i];
    for (j = 0; j < (int)s[i].size(); ++j) {
      int t = 0;
      for (int k = j; k < (int)s[i].size(); ++k) {
        if (k - j + 1 >= 30) break;
        int l = k - j + 1;
        if (s[i][k] == '1') t |= (1 << (l - 1));
        S[i][l].insert(t);
      }
    }
  }
  scanf("%d", &m);
  for (i = n + 1; i <= n + m; ++i) {
    scanf("%d%d", &x, &y);
    L[i] = R[i] = "";
    if (L[x].size() < 30) {
      string tmp = L[x] + L[y];
      for (j = 0; j < 30 && j < (int)tmp.size(); ++j) L[i] += tmp[j];
    } else
      L[i] = L[x];
    if (R[y].size() < 30) {
      string tmp = R[x] + R[y];
      for (j = tmp.size() - 1; j >= 0 && j >= (int)tmp.size() - 30; --j)
        R[i] = tmp[j] + R[i];
    } else
      R[i] = R[y];
    string tmp = R[x] + L[y];
    for (j = 1; j <= 30; ++j) {
      S[i][j] = S[y][j];
      for (set<int>::iterator it = S[x][j].begin(); it != S[x][j].end(); ++it)
        S[i][j].insert(*it);
    }
    for (int j = 0; j < (int)tmp.size(); ++j) {
      int t = 0;
      for (int k = j; k < (int)tmp.size(); ++k) {
        if (k - j + 1 >= 30) break;
        int l = k - j + 1;
        if (tmp[k] == '1') t |= (1 << (l - 1));
        S[i][l].insert(t);
      }
    }
    int ans = 0;
    for (int j = 1; j <= 30; ++j)
      if ((int)S[i][j].size() == (1 << j)) ans = j;
    printf("%d\n", ans);
  }
  return 0;
}
