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;
vector<string> src;
vector<string> sp, ss;
int n;
set<string> st[201][16];
int proc;
void update(int a, int b) {
  for (int k = 1; k <= 15; ++k)
    for (auto& str : st[a][k]) st[proc][k].insert(str);
  for (int k = 1; k <= 15; ++k)
    for (auto& str : st[b][k]) st[proc][k].insert(str);
  string fix1 = sp[a] + sp[b];
  string fix2 = ss[a] + ss[b];
  int len1 = fix1.length();
  int len2 = fix2.length();
  sp.push_back(fix1.substr(0, min(len1, 15)));
  ss.push_back(fix2.substr(max(0, len2 - 15), min(len2, 15)));
  string con = ss[a] + sp[b];
  for (int k = 1; k <= 15; ++k)
    for (int i = 0; i + k <= con.length(); ++i)
      st[proc][k].insert(con.substr(i, k));
}
int solve() {
  for (int i = 15; i >= 1; --i)
    if (st[proc][i].size() == (1 << i)) return i;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  src.push_back("");
  sp.push_back("");
  ss.push_back("");
  cin >> n;
  string tmp;
  int len;
  for (int i = 1; i <= n; ++i) {
    cin >> tmp;
    src.push_back(tmp);
    len = tmp.length();
    sp.push_back(tmp.substr(0, min(len, 15)));
    ss.push_back(tmp.substr(max(0, len - 15), min(len, 15)));
  }
  for (int x = 1; x <= n; ++x) {
    len = src[x].length();
    for (int k = 1; k <= 15; ++k)
      for (int i = 0; i + k <= len; ++i) st[x][k].insert(src[x].substr(i, k));
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    proc = n + i;
    int a, b;
    cin >> a >> b;
    update(a, b);
    cout << solve() << '\n';
  }
  return 0;
}