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>
inline int pidorand() { return ((rand() & 32767) << 15) | (rand() & 32767); }
using namespace std;
inline int nxt() {
  int x;
  scanf("%d", &x);
  return x;
}
const int K = 18;
struct String {
  array<bitset<(1 << K)>, K> a;
  string start, finish;
  String() { a[0][0] = 1; }
  String(const string& s) {
    a[0][0] = 1;
    for (int l = 1; l <= (int)s.length() && l < K; ++l) {
      int cur = 0;
      for (int i = 0; i < (int)s.length(); ++i) {
        cur = (cur * 2 + (s[i] - '0')) & ((1 << l) - 1);
        if (i >= l - 1) {
          a[l][cur] = 1;
        }
      }
    }
    start = s.substr(0, min((int)s.length(), K));
    finish = s.substr((int)s.length() - min((int)s.length(), K));
  }
  int get() {
    for (int i = 0; i < K; ++i) {
      if ((int)a[i].count() < (1 << i)) {
        return i - 1;
      }
    }
    return K - 1;
  }
};
String merge(const String& fs, const String& sn) {
  String res(fs.finish + sn.start);
  if (fs.start.length() >= K) {
    res.start = fs.start;
  }
  if (sn.finish.length() >= K) {
    res.finish = sn.finish;
  }
  for (int i = 0; i < K; ++i) {
    res.a[i] |= fs.a[i] | sn.a[i];
  }
  return res;
}
int main() {
  int n = nxt();
  vector<String> a(n);
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    a[i] = String(s);
  }
  int m = nxt();
  a.reserve(n + m);
  while (m--) {
    int i = nxt() - 1, j = nxt() - 1;
    a.push_back(merge(a[i], a[j]));
    printf("%d\n", a.back().get());
  }
  return 0;
}