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 MAXL = (1 << 15), MAXN = 505;
string s[MAXN], pre[MAXN], suf[MAXN];
int Ans[MAXN], Len[MAXN];
bool H[MAXL];
bool Check(string s, int k) {
  int cnt = 0;
  for (int i = k - 1; i < s.length(); i++) {
    int tmp = 0;
    for (int j = i - k + 1; j <= i; j++) tmp = (tmp << 1) | (s[j] == '1');
    if (i >= k - 1) {
      if (!H[tmp]) cnt++;
      H[tmp] = 1;
    }
  }
  for (int i = 0; i < s.length(); i++) {
    int tmp = 0;
    for (int j = i - k + 1; j <= i; j++) tmp = (tmp << 1) | (s[j] == '1');
    if (i >= k - 1) H[tmp] = 0;
  }
  if ((1 << k) == cnt) return true;
  return false;
}
int Cal(string s) {
  int Ans = 1;
  while (true) {
    if (!Check(s, Ans)) return Ans - 1;
    Ans++;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    Ans[i] = Cal(s[i]);
    pre[i] = suf[i] = s[i];
    Len[i] = s[i].length();
  }
  int m;
  cin >> m;
  int a, b;
  for (int i = n + 1; i <= n + m; i++) {
    cin >> a >> b;
    if (Len[a] + Len[b] > MAXL) {
      string x;
      x = suf[a] + pre[b];
      Ans[i] = Cal(x);
      pre[i] = pre[a];
      suf[i] = suf[b];
      Len[i] = x.length();
      Ans[i] = max(Ans[i], max(Ans[a], Ans[b]));
    } else {
      s[i] = pre[i] = suf[i] = s[a] + s[b];
      Len[i] = Len[a] + Len[b];
      Ans[i] = Cal(s[i]);
    }
    cout << Ans[i] << endl;
  }
  return 0;
}