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;
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, ans[205];
string s[205];
int work(int x) {
  for (int i = 1; i <= 10; ++i)
    for (int j = 0; j <= (1 << i) - 1; ++j) {
      string t;
      for (int k = 0; k <= i - 1; ++k) {
        if (j & (1 << k))
          t += '1';
        else
          t += '0';
      }
      if (s[x].find(t) == -1) return i - 1;
    }
  return 10;
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) cin >> s[i];
  m = read();
  for (int i = 1; i <= n; ++i) ans[i] = work(i);
  for (int i = 1; i <= m; ++i) {
    int l = read(), r = read();
    s[n + i] = s[l] + s[r];
    ans[n + i] = max(work(n + i), max(ans[l], ans[r]));
    if (s[n + i].length() > 512)
      s[n + i] = s[n + i].substr(0, 256) +
                 s[n + i].substr(s[n + i].length() - 256, 256);
    printf("%d\n", ans[i + n]);
  }
}