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 long long oo = 1000000000000000000;
const int N = 205;
set<string> v[N][10];
char second[N];
int mx[N];
string s[N];
int n;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < int(n); i++) {
    scanf(" %s", second);
    s[i] = second;
    for (int sz = 1; sz <= int(9); sz++) {
      if (s[i].size() < sz) break;
      for (int j = 0; j < int(s[i].size() - sz + 1); j++) {
        v[i][sz].insert(s[i].substr(j, sz));
      }
      if (v[i][sz].size() == (1 << sz)) mx[i] = sz;
    }
  }
  int m;
  scanf("%d", &m);
  int a, b;
  for (int i = int(n); i <= int(n + m - 1); i++) {
    scanf("%d %d", &a, &b);
    a--, b--;
    mx[i] = max(mx[a], mx[b]);
    for (int j = 0; j < int(10); j++) {
      for (const string &x : v[a][j]) v[i][j].insert(x);
      for (const string &x : v[b][j]) v[i][j].insert(x);
      if (v[i][j].size() == (1 << j)) mx[i] = max(mx[i], j);
    }
    for (int sz = int(2); sz <= int(9); sz++) {
      for (int j = 1; j <= int(sz - 1); j++) {
        if (s[b].size() < j) break;
        if (s[a].size() < sz - j) continue;
        v[i][sz].insert(s[a].substr(s[a].size() - (sz - j), sz - j) +
                        s[b].substr(0, j));
      }
      if (v[i][sz].size() == (1 << sz)) mx[i] = max(mx[i], sz);
    }
    printf("%d\n", mx[i]);
    s[i] = s[a] + s[b];
    if (s[i].size() > 20)
      s[i] = s[i].substr(0, 10) + s[i].substr(s[i].size() - 10, 10);
  }
  return 0;
}