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>
#pragma comment(linker, "/STACK:167772160000")
using namespace std;
struct str {
  string beg;
  string end;
  set<int> contains[30];
};
str unite(str a, str b) {
  str ret;
  string newString = a.end + b.beg;
  for (int i = 0; i < (15); i++) {
    if (a.contains[i].size() > b.contains[i].size()) {
      ret.contains[i] = a.contains[i];
      for (auto x : b.contains[i]) ret.contains[i].insert(x);
    } else {
      ret.contains[i] = b.contains[i];
      for (auto x : a.contains[i]) ret.contains[i].insert(x);
    }
  }
  for (int i = 0; i < newString.size(); i++)
    for (int j = i; j < newString.size(); j++) {
      if (j - i + 1 > 15) continue;
      int curString = 0;
      for (int k = i; k <= j; k++)
        if (newString[k] == '1') curString |= (1 << (k - i));
      ret.contains[j - i].insert(curString);
    }
  if (a.beg.length() < 15) {
    ret.beg = newString;
    if (ret.beg.length() > 15) ret.beg = ret.beg.substr(0, 15);
  } else
    ret.beg = a.beg;
  if (b.end.length() < 15) {
    ret.end = newString;
    if (ret.end.length() > 15)
      ret.end = ret.end.substr(ret.end.length() - 15, 15);
  } else
    ret.end = b.end;
  return ret;
}
str strings[300];
str zer;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= (n); i++) {
    str s = zer;
    cin >> s.beg;
    s.end = s.beg;
    strings[i] = unite(s, zer);
  }
  int m;
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    int a, b;
    cin >> a >> b;
    strings[i] = unite(strings[a], strings[b]);
    int j = 0;
    while (strings[i].contains[j].size() == (1 << (j + 1))) j++;
    cout << j << endl;
  }
  return 0;
}