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;
int const N = 205;
int const NN = 11;
int ans[N], n;
bitset<1 << NN> a[N][NN];
string last[N], first[N], s[N];
void add(int index, string s) {
  for (int i = 0; i < s.size(); i++) {
    int val = 0;
    for (int j = i; j < min((int)s.size(), i + NN); j++) {
      val = val * 2 + (s[j] - '0');
      a[index][j - i + 1][val] = 1;
    }
  }
  while (1) {
    for (int mask = 0; mask < (1 << (ans[index] + 1)); mask++) {
      if (a[index][ans[index] + 1][mask] == 0) {
        return;
      }
    }
    ans[index]++;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    if (s[i].size() < NN) {
      last[i] = first[i] = s[i];
    } else {
      last[i] = s[i].substr(s[i].size() - NN, NN);
      first[i] = s[i].substr(0, NN);
    }
    add(i, s[i]);
  }
  int m;
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    int u, v;
    cin >> u >> v;
    if (last[v].size() < NN) {
      int rest = min(NN - last[v].size(), last[u].size());
      last[i] = last[u].substr(last[u].size() - rest, rest) + last[v];
    } else {
      last[i] = last[v];
    }
    if (first[u].size() < NN) {
      int rest = min(NN - first[u].size(), first[v].size());
      first[i] = first[u] + first[v].substr(0, rest);
    } else {
      first[i] = first[u];
    }
    for (int j = 1; j < NN; j++) a[i][j] = a[u][j] | a[v][j];
    ans[i] = max(ans[u], ans[v]);
    string newStr = last[u] + first[v];
    add(i, newStr);
    cout << ans[i] << endl;
  }
}
