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>
#pragma comment(linker, "/STACK:1000000000")
using namespace std;
const int maxn = 210, magic = 15;
string s[maxn];
bitset<(1 << magic)> b[maxn][magic + 1];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    for (int j = 0; j < (int)s[i].size(); j++) {
      int now = 0;
      for (int k = j; k < (int)s[i].size(); k++) {
        if (k - j + 1 > magic) {
          break;
        }
        now <<= 1;
        now += s[i][k] - '0';
        b[i][k - j + 1][now] = 1;
      }
    }
  }
  int m;
  cin >> m;
  int p = n;
  while (m--) {
    int a, c;
    cin >> a >> c;
    a--;
    c--;
    s[p] = s[a] + s[c];
    for (int i = 1; i <= magic; i++) {
      b[p][i] = b[a][i] | b[c][i];
    }
    for (int i = 1; i <= (int)s[a].size(); i++) {
      if (i > magic) {
        break;
      }
      int now = 0;
      for (int j = (int)s[a].size() - i; j < (int)s[a].size(); j++) {
        now <<= 1;
        now += s[a][j] - '0';
      }
      for (int j = 1; j <= (int)s[c].size(); j++) {
        if (i + j > magic) {
          break;
        }
        now <<= 1;
        now += s[c][j - 1] - '0';
        b[p][i + j][now] = 1;
      }
    }
    int ans = 0;
    for (int i = 1; i <= magic; i++) {
      int fl = 0;
      for (int j = 0; j < (1 << i); j++) {
        if (b[p][i][j] == 0) {
          fl = 1;
          break;
        }
      }
      if (fl == 0) {
        ans++;
      } else {
        break;
      }
    }
    if ((int)s[p].size() > 2 * magic) {
      string s1, s2;
      for (int i = 0; i < magic; i++) {
        s1.push_back(s[p][i]);
      }
      for (int i = (int)s[p].size() - magic; i < (int)s[p].size(); i++) {
        s2.push_back(s[p][i]);
      }
      s[p] = s1 + s2;
    }
    p++;
    cout << ans << endl;
  }
  return 0;
}