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 long long MAXN = 333, K = 15, MOD = 1e9 + 7;
string pre[333], suf[333];
set<long long> t[MAXN][K];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    string s;
    cin >> s;
    long long m = s.size();
    for (long long i = 0; i < m; i++) s[i] -= '0';
    for (long long j = 0; j < m; j++) {
      long long val = 0;
      for (long long k = j; k < m; k++)
        if (k - j + 1 < K) val = 2 * val + s[k], t[i][k - j + 1].insert(val);
    }
    for (long long w = 0; w < min(m, K); w++)
      pre[i] += s[w], suf[i] += s[m - w - 1];
    reverse(suf[i].begin(), suf[i].end());
    t[i][0].insert(0);
  }
  long long m;
  cin >> m;
  for (long long i = n; i < n + m; i++) {
    long long id = i;
    long long x, y;
    cin >> x >> y;
    x--, y--;
    string q = pre[x] + pre[y], w = suf[x] + suf[y];
    pre[id] = suf[id] = "";
    long long k = q.size();
    for (long long i = 0; i < min(K, k); i++) pre[id] += q[i];
    k = w.size();
    for (long long i = 0; i < min(K, k); i++) suf[id] += w[k - i - 1];
    reverse(suf[i].begin(), suf[i].end());
    for (long long w = 0; w < K; w++) {
      for (auto u : t[x][w]) t[i][w].insert(u);
      for (auto u : t[y][w]) t[i][w].insert(u);
    }
    long long l = suf[x].size(), r = pre[y].size();
    long long v = 0;
    id = i;
    for (long long i = l - 1; i > -1; i--) {
      long long v = 0;
      for (long long j = i; j < l; j++) v = 2 * v + suf[x][j];
      long long val = v;
      for (long long j = 0; j < r; j++) {
        val = 2 * val + pre[y][j];
        if (j + 1 + l - i < K) t[id][j + 1 + l - i].insert(val);
      }
    }
    long long ans = 1;
    while (ans < K && t[i][ans].size() == (1 << ans)) ans++;
    cout << ans - 1 << '\n';
  }
  return 0;
}