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[222], str;
int res[1000000] = {0};
bool hh(int n, int x) {
  for (int i = 0; i < (1 << x); i++) {
    str = "";
    for (int k = 0; k < x; k++) (i & (1 << k)) ? str += "1" : str += "0";
    if (s[n].find(str) == -1) return false;
  }
  return true;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    int ii = 0, jj = 10;
    while (ii < jj && hh(i, ii + 1)) ii++;
    res[i] = ii;
  }
  int t, m;
  cin >> t;
  while (t--) {
    int i, j, mid, ii, jj;
    cin >> i >> j;
    ii = i;
    jj = j;
    n++;
    s[n] = s[i] + s[j];
    if (s[n].size() > 1000)
      s[n] = s[n].substr(0, 500) + s[n].substr(s[n].size() - 500, 500);
    i = 0;
    j = 10;
    while (i < j && hh(n, i + 1)) i++;
    res[n] = max(i, max(res[ii], res[jj]));
    cout << res[n] << "\n";
  }
  return 0;
}
