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 maxn = 1e6 + 10;
const int INF = 0x3f3f3f3f;
string str[210], s;
int Hash[maxn];
pair<int, int> to[210];
int n, a, b;
int ans[210];
int main() {
  int DEG = 15;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> str[i];
  }
  int q;
  cin >> q;
  int dfn = 0;
  int tot = n + 1;
  while (q--) {
    cin >> a >> b;
    to[tot].first = a, to[tot].second = b;
    str[tot] = str[a] + str[b];
    s = str[tot];
    for (int i = 1; i < DEG; i++) {
      dfn++;
      int x = 0;
      if (i > s.length()) {
        ans[tot] = max(ans[a], ans[b]);
        ans[tot] = max(ans[tot], i - 1);
        cout << ans[tot] << endl;
        break;
      }
      for (int k = 0; k < i; k++) {
        x <<= 1;
        x |= s[k] - '0';
      }
      Hash[x] = dfn;
      for (int j = i; j < s.length(); j++) {
        x &= (1 << (i - 1)) - 1;
        x <<= 1;
        x |= s[j] - '0';
        Hash[x] = dfn;
      }
      int flag = 1;
      for (int j = 0; j < 1 << i; j++) {
        if (Hash[j] != dfn) {
          flag = 0;
          break;
        }
      }
      if (flag == 0) {
        ans[tot] = max(ans[a], ans[b]);
        ans[tot] = max(ans[tot], i - 1);
        cout << ans[tot] << endl;
        break;
      }
    }
    if (str[tot].length() > 2000) {
      str[tot] = str[tot].substr(0, 1000) +
                 str[tot].substr(str[tot].length() - 1000, str[tot].length());
    }
    tot++;
  }
  return 0;
}