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.
#include <bits/stdc++.h>
bool f(const std::string &s, long long k) {
  if (k > s.size()) return false;
  long long bits = 0;
  std::set<long long> strs;
  for (int i = 0; i < k; ++i) {
    bits = (bits << 1) | (s[i] - '0');
  }
  strs.insert(bits);
  for (int i = k; i < s.size(); ++i) {
    bits = (bits << 1) | (s[i] - '0');
    strs.insert(bits & ((1LL << k) - 1));
  }
  return strs.size() == (1LL << k);
}
int n, m;
std::string strs[205];
int ans[205] = {0};
void f(int i) {
  while (f(strs[i], ans[i] + 1)) ++ans[i];
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  std::cin >> n;
  std::vector<std::string> vs(n);
  for (int i = 1; i <= n; ++i) {
    std::cin >> strs[i];
    f(i);
  }
  std::cin >> m;
  for (int i = 1; i <= m; ++i) {
    int a, b;
    std::cin >> a >> b;
    strs[n + i] = strs[a] + strs[b];
    ans[n + i] = std::max(ans[a], ans[b]);
    f(n + i);
    std::cout << ans[n + i] << '\n';
    if (strs[n + i].size() > 1e6)
      strs[n + i] = strs[n + i].substr(0, 1e5) +
                    strs[n + i].substr(strs[n + i].size() - 1e5);
  }
  return 0;
}
