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 MAXANS = 11;
vector<set<string>> count(string s) {
  vector<set<string>> a(MAXANS + 1);
  for (int k = 1; k <= min(int(s.length()), MAXANS); ++k) {
    set<string> x;
    for (int i = 0; i < s.length() - k + 1; ++i) {
      x.insert(s.substr(i, k));
    }
    a[k] = x;
  }
  return a;
}
class Stringg {
 public:
  string begin;
  string end;
  vector<set<string>> substr;
  Stringg(const string& s) {
    begin = s.substr(0, max(MAXANS, int(s.length())));
    end = s.substr(max(0, int(s.length() - MAXANS)));
    substr = count(s);
  }
  Stringg(const Stringg& a, const Stringg& b) {
    begin = a.begin;
    if (begin.length() < MAXANS) {
      int k = min(MAXANS - begin.length(), b.begin.length());
      begin += b.begin.substr(0, k);
    }
    if (b.end.length() >= MAXANS) {
      end = b.end;
    } else {
      int k = min(MAXANS - b.end.length(), a.end.length());
      end = a.end.substr(a.end.length() - k) + b.end;
    }
    substr = count(a.end + b.begin);
    for (int k = 1; k <= MAXANS; ++k) {
      for (auto x : a.substr[k]) {
        substr[k].insert(x);
      }
      for (auto x : b.substr[k]) {
        substr[k].insert(x);
      }
    }
  }
  int getK() {
    long long l = 2;
    long long ans = 0;
    for (int k = 1; k <= MAXANS; ++k) {
      if (substr[k].size() == l) {
        ans = k;
      }
      l *= 2;
    }
    return ans;
  }
};
int main() {
  int n;
  cin >> n;
  vector<vector<set<string>>> a;
  vector<Stringg> b;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    b.push_back(Stringg(s));
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; ++i) {
    int k, l;
    cin >> k >> l;
    b.push_back(Stringg(b[k - 1], b[l - 1]));
    cout << b[n + i].getK() << endl;
  }
  return 0;
}