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;
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& p) {
  return o << "(" << p.first << "," << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& vc) {
  o << "sz = " << vc.size() << endl << "[";
  for (const T& v : vc) o << v << ",";
  o << "]";
  return o;
}
int bsr(int x) {
  if (x == 0) return -1;
  return 31 ^ __builtin_clz(x);
}
bool B(int x, int i) { return (x >> i) & 1; }
int N;
const int X = 12;
const int MX = 1 << (X + 1);
using BS = bitset<MX>;
BS has[1000];
string le[1000], ri[1000];
int sz(int x) { return le[x].size(); }
BS calchas(string s) {
  BS res;
  int N = s.size();
  int I = 2;
  for (int len = 1; len <= (int)(X); len++)
    for (int b = 0; b < (int)(1 << len); b++) {
      string t;
      for (int i = 0; i < (int)(len); i++) t += '0' + B(b, i);
      int M = t.size();
      bool ok = 0;
      for (int i = 0; i < (int)(N - M + 1); i++) {
        if (s.substr(i, M) == t) {
          ok = 1;
          break;
        }
      }
      res[I] = ok;
      I++;
    }
  return res;
}
int main() {
  cin >> N;
  for (int i = 0; i < (int)(N); i++) {
    string s;
    cin >> s;
    le[i] = s;
    ri[i] = s;
    has[i] = calchas(s);
  }
  int Q;
  cin >> Q;
  for (int i = 0; i < (int)(Q); i++) {
    int id = N + i;
    int x, y;
    cin >> x >> y;
    x--, y--;
    {
      string t = le[x] + le[y];
      le[id] = t.substr(0, X);
    }
    {
      string t = ri[x] + ri[y];
      if (t.size() <= X) {
        ri[id] = t;
      } else {
        ri[id] = t.substr(t.size() - X, X);
      }
    }
    {
      string t = ri[x] + le[y];
      has[id] = has[x] | has[y] | calchas(t);
    }
    bool done = 0;
    for (int i = 2; i < MX; i++)
      if (!has[id][i]) {
        cout << bsr(i) - 1 << endl;
        done = 1;
        break;
      }
    if (!done) {
      cout << X << endl;
    }
  }
}