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;
constexpr int sz = 11;
constexpr int ssz = (1 << sz);
unordered_set<string> mp[sz];
constexpr int nn = 205;
string a[nn];
int dp[nn];
void prepare() {
  for (int i = 1; i < sz; ++i) {
    for (int j = 0; j < (1 << i); ++j) {
      string cur = "";
      for (int k = 0; k < i; ++k) {
        if ((1 << k) & j)
          cur += "1";
        else
          cur += "0";
      }
      mp[i].insert(cur);
    }
  }
}
int solve(const string& s) {
  for (int i = 0; i < sz; ++i) {
    for (auto t : mp[i + 1]) {
      if (s.find(t) == string::npos) return i;
    }
  }
}
int main() {
  prepare();
  int n, m;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    dp[i] = solve(a[i]);
  }
  cin >> m;
  for (int i = n + 1, x, y; i <= n + m; ++i) {
    cin >> x >> y;
    a[i] = a[x] + a[y];
    if (a[i].length() > ssz)
      a[i] = a[i].substr(0, ssz / 2) +
             a[i].substr(a[i].length() - ssz / 2, ssz / 2);
    dp[i] = max(dp[x], max(dp[y], solve(a[i])));
    cout << dp[i] << endl;
  }
}