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>
using namespace std;
const int N = 111;
const int L = 10;
struct word {
  string prefix, suffix;
  bool seen[L + 1][1 << L];
};
void add(const string &s, bool seen[][1 << L]) {
  int n = ((int)(s).size());
  for (int l = (int)(1); l < ((int)L + 1); l++) {
    int mask = 0;
    for (int i = 0; i < (int)(n); i++) {
      mask = 2 * mask + s[i] - '0';
      if (i >= l) mask -= int(s[i - l] - '0') << l;
      if (i >= l - 1) seen[l][mask] = 1;
    }
  }
}
string prefix(const string &s) {
  return s.substr(0, min(((int)(s).size()), L));
}
string suffix(string s) {
  reverse((s).begin(), (s).end());
  s = prefix(s);
  reverse((s).begin(), (s).end());
  return s;
}
word combine(const word &a, const word &b) {
  word ans;
  for (int l = (int)(1); l < ((int)L + 1); l++)
    for (int mask = 0; mask < (int)(1 << l); mask++)
      ans.seen[l][mask] = a.seen[l][mask] || b.seen[l][mask];
  add(a.suffix + b.prefix, ans.seen);
  ans.prefix = prefix(a.prefix + b.prefix);
  ans.suffix = suffix(a.suffix + b.suffix);
  return ans;
}
word build(const string &s) {
  word ans;
  memset(ans.seen, 0, sizeof ans.seen);
  add(s, ans.seen);
  ans.prefix = prefix(s);
  ans.suffix = suffix(s);
  return ans;
}
int go(const word &w) {
  for (int l = (int)(1); l < ((int)L + 1); l++)
    for (int mask = 0; mask < (int)(1 << l); mask++)
      if (!w.seen[l][mask]) return l - 1;
}
int n;
word w[2 * N];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n); i++) {
    string s;
    cin >> s;
    w[i] = build(s);
  }
  int q;
  cin >> q;
  for (int i = (int)(n); i < ((int)n + q); i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    w[i] = combine(w[x], w[y]);
    cout << go(w[i]) << '\n';
  }
  return 0;
}
