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;
const int MAX = 200020;
const long long MOD = (int)1e9 + 7;
const int INF = 1e9;
const long long LLINF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-7;
void get_sub(string s, int size, set<string>& sub) {
  int n = s.size();
  for (int l = 0, r = size - 1; r < n; l++, r++) {
    string t = s.substr(l, size);
    sub.insert(t);
  }
}
set<string> merge(set<string> a, set<string> b) {
  set<string> res = a;
  for (auto s : b) {
    res.insert(s);
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, m;
  cin >> n;
  vector<string> pre(301, string()), suf(301, string());
  vector<vector<set<string>>> sub(301, vector<set<string>>(10, set<string>()));
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    int sz = s.size();
    pre[i] = s.substr(0, min(9, sz));
    suf[i] = s.substr(max(0, sz - 9), min(9, sz));
    for (int k = 1; k <= 9; k++) {
      get_sub(s, k, sub[i][k]);
    }
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    string t = suf[a] + pre[b];
    int sz = t.size();
    string p = pre[a] + pre[b];
    sz = p.size();
    pre[n + i] = p.substr(0, min(9, sz));
    string s = suf[a] + suf[b];
    sz = s.size();
    suf[n + i] = s.substr(max(0, sz - 9), min(9, sz));
    for (int k = 1; k <= 9; k++) {
      sub[n + i][k] = merge(sub[a][k], sub[b][k]);
      get_sub(t, k, sub[n + i][k]);
    }
    int res = 0;
    for (int k = 1; k <= 9; k++) {
      if ((int)sub[n + i][k].size() == (1 << k)) res = k;
    }
    cout << res << '\n';
  }
  return 0;
}