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;
set<string> st[303][12];
int n;
string s[303];
string suf[303];
string pre[303];
long long len[303];
void combine(int idx, int i, int j) {
  for (int len = 1; len < 12; len++) {
    for (auto it = st[i][len].begin(); it != st[i][len].end(); it++) {
      st[idx][len].insert(*it);
    }
    for (auto it = st[j][len].begin(); it != st[j][len].end(); it++) {
      st[idx][len].insert(*it);
    }
  }
  string comb = suf[i] + pre[j];
  for (int k = 0; k < comb.size(); k++) {
    string b = "";
    for (int d = k; d < min((int)comb.size(), k + 11); d++) {
      b += comb[d];
      st[idx][b.size()].insert(b);
    }
  }
  suf[idx] = suf[j];
  if (len[j] < 11) {
    for (int k = suf[i].size() - 1; k >= 0; k--) {
      suf[idx].insert(suf[idx].begin(), suf[i][k]);
      if (suf[idx].size() == 11) break;
    }
  }
  pre[idx] = pre[i];
  if (len[i] < 11) {
    for (int k = 0; k < pre[j].size(); k++) {
      pre[idx].push_back(pre[j][k]);
      if (pre[idx].size() == 11) break;
    }
  }
  len[idx] = min((long long)40, len[i] + len[j]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    len[i] = s[i].size();
    pre[i] = s[i];
    suf[i] = s[i];
    for (int j = 0; j < s[i].length(); j++) {
      string b = "";
      for (int k = j; k < min((int)s[i].length(), j + 11); k++) {
        b += s[i][k];
        st[i][b.size()].insert(b);
      }
    }
  }
  int q;
  cin >> q;
  int cur = n;
  for (int i = 1; i <= q; i++) {
    int l;
    int r;
    cin >> l >> r;
    combine(++cur, l, r);
    int ans = 0;
    for (int j = 11; j >= 0; j--) {
      if (st[cur][j].size() == (1 << j)) {
        ans = j;
        break;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
