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 size_t MAXLEN = 12;
string a[205];
set<pair<int, int>> ima[205];
int n, m;
void nadji(set<pair<int, int>>& ima, const string& s) {
  for (size_t i = 0; i < s.size(); i++) {
    int mask = 0;
    for (size_t j = 1; j <= MAXLEN; j++) {
      if (i + j - 1 >= s.size()) {
        break;
      }
      mask = 2 * mask + s[i + j - 1] - '0';
      ima.insert({j, mask});
    }
  }
}
void nadji(int id) { nadji(ima[id], a[id]); }
void spoji(int id, int x, int y) {
  a[id] = a[x] + a[y];
  if (a[id].size() > 2 * MAXLEN) {
    a[id] =
        a[id].substr(0, MAXLEN) + a[id].substr(a[id].size() - MAXLEN, MAXLEN);
  }
  ima[id] = ima[x];
  for (auto p : ima[y]) {
    ima[id].insert(p);
  }
  string between =
      a[x].size() > MAXLEN ? a[x].substr(a[x].size() - MAXLEN, MAXLEN) : a[x];
  between += a[y].size() > MAXLEN ? a[y].substr(0, MAXLEN) : a[y];
  nadji(ima[id], between);
}
int broj(int id) {
  for (int k = MAXLEN; k > 0; k--) {
    bool ok = true;
    for (int mask = 0; mask < (1 << k); mask++) {
      if (!ima[id].count({k, mask})) {
        ok = false;
        break;
      }
    }
    if (ok) {
      return k;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    nadji(i);
  }
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    int x, y;
    cin >> x >> y;
    spoji(i, x, y);
    cout << broj(i) << '\n';
  }
}