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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res;
}
long long raichu(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = ((res) * (x));
    y = y >> 1;
    x = ((x) * (x));
  }
  return res;
}
bool isprime(long long n) {
  if (n < 2)
    return false;
  else if (n == 2)
    return true;
  else if (n % 2 == 0)
    return false;
  else {
    long long z = sqrt(n);
    for (int i = 0; i < z - 1; i++)
      if (n % (i + 2) == 0) return false;
    return true;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tests = 1, Tests;
  Tests = tests;
  while (tests--) {
    int n;
    cin >> n;
    vector<pair<string, string> > s(n);
    int MA = 10;
    vector<vector<set<string> > > v(n, vector<set<string> >(MA + 1));
    for (int i = 0; i < n; i++) {
      cin >> s[i].first;
      s[i].second = s[i].first;
      for (int j = 1; j <= MA; j++) {
        for (int k = 0; k < (int)s[i].second.size() - j + 1; k++) {
          string z = s[i].second.substr(k, j);
          v[i][j].insert(z);
        }
      }
    }
    int m;
    cin >> m;
    for (int i = 0; i < m; i++) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      string X, Y;
      if ((int)s[x].first.size() < 100) {
        X = s[x].first;
        for (int j = 0; j < (int)s[y].first.size(); j++) {
          X.push_back(s[y].first[j]);
          if ((int)X.size() == 100) break;
        }
      } else {
        X = s[x].first;
      }
      if ((int)s[y].second.size() < 100) {
        Y = s[y].second;
        reverse(Y.begin(), Y.end());
        for (int j = 0; j < (int)s[x].second.size(); j++) {
          Y.push_back(s[x].second[(int)s[x].second.size() - 1 - j]);
          if ((int)Y.size() == 100) break;
        }
        reverse(Y.begin(), Y.end());
      } else {
        Y = s[y].second;
      }
      s.push_back(make_pair(X, Y));
      vector<set<string> > V(MA + 1);
      for (int j = 1; j <= MA; j++) {
        V[j] = v[y][j];
        for (auto k : v[x][j]) {
          V[j].insert(k);
        }
      }
      v.push_back(V);
      for (int j = 1; j <= MA; j++) {
        for (int k = 0; k < (int)s[(int)s.size() - 1].second.size() - j + 1;
             k++) {
          string z = s[(int)s.size() - 1].second.substr(k, j);
          v[(int)v.size() - 1][j].insert(z);
        }
        for (int k = 0; k < (int)s[(int)s.size() - 1].first.size() - j + 1;
             k++) {
          string z = s[(int)s.size() - 1].first.substr(k, j);
          v[(int)v.size() - 1][j].insert(z);
        }
      }
      int ans = 0;
      for (int j = 1; j <= MA; j++) {
        if ((int)v[(int)v.size() - 1][j].size() == (1LL << j)) {
          ans = j;
        }
      }
      cout << ans << endl;
    }
  }
  return 0;
}