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 N = 300031;
const int MAG = 11;
set<string> S[300][MAG + 2];
set<string>::iterator it;
int m;
string pref[N], suf[N], st[N];
int n;
string get_pref(string st) {
  while (st.size() > MAG) st.erase(st.end() - 1);
  return st;
}
string get_suf(string st) {
  while (st.size() > MAG) st.erase(st.begin());
  return st;
}
void run_merge(int a, int b, int to) {
  for (int i = 1; i < MAG; i++) {
    for (it = S[a][i].begin(); it != S[a][i].end(); it++) {
      string here = (*it);
      S[to][i].insert(here);
    }
    for (it = S[b][i].begin(); it != S[b][i].end(); it++) {
      string here = (*it);
      S[to][i].insert(here);
    }
  }
}
void process_string(string st, int id) {
  for (int i = 0; i < st.size(); i++) {
    string temp = "";
    for (int j = i; j < st.size(); j++) {
      temp += st[j];
      if (temp.size() >= MAG) break;
      S[id][temp.size()].insert(temp);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> st[i];
    pref[i] = get_pref(st[i]);
    suf[i] = get_suf(st[i]);
    process_string(st[i], i);
  }
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    int a, b;
    cin >> a >> b;
    run_merge(a, b, i);
    string here = suf[a] + pref[b];
    st[i] = here;
    process_string(st[i], i);
    pref[i] = pref[a] + pref[b];
    suf[i] = suf[a] + suf[b];
    pref[i] = get_pref(pref[i]);
    suf[i] = get_suf(suf[i]);
    int ans = 1;
    while (S[i][ans].size() == (1ll << ans)) ++ans;
    cout << ans - 1 << endl;
  }
  cin.get();
  cin.get();
  return 0;
}