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;
typedef class String {
 public:
  string pre, suf;
  bool overflow;
  int rt;
  bitset<65536> mark;
  String() : overflow(false) {}
} String;
String operator+(String& a, String& b) {
  String rt;
  rt.overflow = a.overflow || b.overflow;
  rt.mark = a.mark | b.mark;
  if (!a.overflow) {
    rt.pre = a.pre + b.pre;
    if (rt.pre.length() > 15) rt.overflow = true, rt.pre.resize(15);
  } else
    rt.pre = a.pre;
  if (!b.overflow) {
    rt.suf = a.suf + b.suf;
    if (rt.suf.length() > 15)
      rt.overflow = true,
      rt.suf = rt.suf.substr(rt.suf.size() - 15, rt.suf.size());
  } else
    rt.suf = b.suf;
  string s = a.suf + b.pre;
  for (int i = 0; i < s.length(); i++)
    for (int j = 0, t = 0; j < 15 && i + j < s.length(); j++) {
      t = (t << 1) | (s[i + j] & 1);
      rt.mark[t | (1 << (j + 1))] = 1;
    }
  int i;
  for (rt.rt = 0;; rt.rt++) {
    for (i = 0; i < (1 << rt.rt + 1) && rt.mark[i | (1 << rt.rt + 1)]; i++)
      ;
    if (!rt.mark[i | (1 << rt.rt + 1)]) break;
  }
  return rt;
}
int n, m;
String strs[205];
inline void init() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> strs[i].pre;
    strs[i].suf = strs[i].pre;
    strs[i].mark[1] = 1;
    for (int j = 0; j < strs[i].pre.length(); j++)
      for (int k = 0, t = 0; k < 15 && j + k < strs[i].pre.length(); k++) {
        t = (t << 1) | (strs[i].pre[j + k] & 1);
        strs[i].mark[t | (1 << (k + 1))] = 1;
      }
  }
}
inline void solve() {
  cin >> m;
  for (int i = 1, a, b; i <= m; i++) {
    cin >> a >> b;
    strs[n + i] = strs[a] + strs[b];
    cout << strs[n + i].rt << endl;
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  init();
  solve();
  return 0;
}