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;
short bit[21][205][20005], len[205], idd;
string ls[205], rs[205], s[205];
void cross(string b) {
  string a = b;
  reverse(a.begin(), a.end());
  for (int j = 0; j < a.size(); j++) {
    int val = 0;
    for (int k = j; k < a.size() && k < j + 20; k++) {
      if (a[k] == '1') val |= ((1ll << ((k - j))));
      if (val >= 2e4) break;
      bit[k - j + 1][idd][val] = 1;
    }
  }
}
int32_t main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    idd = i;
    len[i] = s[i].size();
    if (s[i].size() >= 20) {
      ls[i] = s[i].substr(0, 20);
      rs[i] = s[i].substr(s[i].size() - 20, 20);
    } else {
      ls[i] = s[i];
      rs[i] = s[i];
    }
    cross(s[i]);
  }
  int m;
  cin >> m;
  for (int i = n + 1; i <= n + m; i++) {
    int l, r;
    cin >> l >> r;
    idd = i;
    len[i] = len[r] + len[l];
    if (len[i] > 20) len[i] = 20;
    for (int k = 1; k <= 20; k++)
      for (int j = 0; j <= 1e4 && j < (1ll << (k)); j++)
        bit[k][i][j] = bit[k][l][j] | bit[k][r][j];
    if (len[l] < 20 && len[r] < 20) {
      s[i] = ls[l] + rs[r];
      len[i] = len[l] + len[r];
      if (s[i].size() >= 20) {
        ls[i] = s[i].substr(0, 20);
        rs[i] = s[i].substr(s[i].size() - 20, 20);
      } else {
        ls[i] = s[i];
        rs[i] = s[i];
      }
      cross(s[i]);
    } else if (len[l] < 20 && len[r] > 19) {
      s[i] = rs[l] + ls[r];
      rs[i] = rs[r];
      ls[i] = s[i].substr(0, 20);
      cross(s[i]);
    } else if (len[i] > 19 && len[r] < 20) {
      s[i] = rs[l] + ls[r];
      ls[i] = ls[l];
      rs[i] = s[i].substr(s[i].size() - 20, 20);
      cross(s[i]);
    } else {
      s[i] = rs[l] + ls[r];
      ls[i] = ls[l];
      rs[i] = rs[r];
      cross(s[i]);
    }
    int id = 0;
    for (int k = 1; k <= 20; k++)
      for (int j = 0; j <= 2e4 && j < (1ll << (k)); j++) {
        if (bit[k][i][j] == 0) break;
        if (j == ((1ll << (k))) - 1) id = k;
      }
    cout << id << "\n";
  }
}