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;
struct edge {
  int v, w;
};
const long long INF = LLONG_MAX / 10;
const int MOD = 1e9 + 7;
void f(string s, set<string>& st) {
  int N = s.length();
  for (int l = 0; l <= N; l++)
    for (int r = l + 1; r <= N; r++)
      if (r - l <= 20) st.insert(s.substr(l, r - l));
}
int main() {
  int N;
  cin >> N;
  vector<string> a(N);
  for (int i = 0; i < N; i++) cin >> a[i];
  vector<set<string> > st(N);
  for (int i = 0; i < N; i++) f(a[i], st[i]);
  int Q;
  cin >> Q;
  while (Q--) {
    int i1, i2;
    cin >> i1 >> i2;
    i1--, i2--;
    string s = a[i1] + a[i2];
    set<string> S;
    for (string s : st[i1]) S.insert(s);
    for (string s : st[i2]) S.insert(s);
    f(s, S);
    if (s.length() > 40) {
      string pre = s.substr(0, 20);
      string suf = s.substr(s.length() - 20);
      s = pre + "#" + suf;
    }
    a.push_back(s), st.push_back(S);
    int k;
    for (k = 1;; k++) {
      bool ok = true;
      for (int T = 0; T < 1 << k; T++) {
        string t(k, ' ');
        for (int j = 0; j < k; j++) t[j] = "01"[T >> j & 1];
        if (!S.count(t)) {
          ok = false;
          break;
        }
      }
      if (!ok) {
        k--;
        break;
      }
    }
    cout << k << endl;
  }
}