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;
int get_id(string s) {
  int sz = s.size(), ans = 0;
  for (int i = 0; i < sz; i++) ans = (ans << 1) | (s[i] & 1);
  return (1 << sz) | ans;
}
struct str {
  bitset<(1 << 17)> bk;
  string pre, suf;
  bool all;
  void update(string s) {
    int up = s.size();
    for (int len = 1; len <= min(up, 16); len++)
      for (int i = 0; i + len <= up; i++) bk[get_id(s.substr(i, len))] = 1;
  }
  void simply() {
    if (pre.size() > 16) pre = pre.substr(0, 16), all = 0;
    if (suf.size() > 16) suf = suf.substr(suf.size() - 16, 16), all = 0;
  }
  int query() {
    for (int i = 16; i >= 0; i--) {
      int ok = 1, up = 1 << i;
      for (int j = 0; j < up; j++)
        if (!bk[up | j]) {
          ok = 0;
          break;
        }
      if (ok) return i;
    }
    return 0;
  }
} ac[210];
void solve(int now, int l, int r) {
  ac[now].bk = ac[l].bk | ac[r].bk;
  if (ac[l].all)
    ac[now].pre = ac[l].pre + ac[r].pre, ac[now].all = 1;
  else
    ac[now].pre = ac[l].pre;
  if (ac[r].all)
    ac[now].suf = ac[l].suf + ac[r].suf, ac[now].all = 1;
  else
    ac[now].suf = ac[r].suf;
  ac[now].update(ac[l].suf + ac[r].pre);
  ac[now].simply();
}
string s;
int main() {
  ios::sync_with_stdio(0);
  int n, m, l, r;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    ac[i].update(s);
    ac[i].pre = ac[i].suf = s;
    ac[i].all = 1;
    ac[i].simply();
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    cin >> l >> r;
    solve(n + i, l, r);
    cout << ac[n + i].query() << endl;
  }
  return 0;
}