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 int K = 13;
string reduce(string a) { return a.substr(0, min(int(a.size()), K)); }
string reduce_reverse(string a) {
  reverse(a.begin(), a.end());
  a = reduce(a);
  reverse(a.begin(), a.end());
  return a;
}
class Info {
 public:
  Info() {
    memset(a, 0, sizeof(a));
    start = "";
    last = "";
  }
  Info(const string& s) {
    memset(a, 0, sizeof(a));
    for (int l = 1; l < K; l++) {
      for (int k = 0; k + l <= s.size(); ++k) {
        int mask = 0;
        for (int it = 0; it < l; it++) {
          mask |= (s[k + it] == '1') << it;
        }
        a[l][mask] = 1;
      }
    }
    start = reduce(s);
    last = reduce_reverse(s);
  }
  string start, last;
  bool a[K][1 << K];
  int ans() {
    for (int i = K - 1; i >= 0; i--) {
      bool flag = true;
      for (int j = 0; j < (1 << i) && flag; ++j) {
        flag &= a[i][j];
      }
      if (flag) return i;
    }
    return false;
  }
  friend Info operator+(const Info& a, const Info& b) {
    Info c;
    for (int l = 1; l < K; l++) {
      for (int mask = 0; mask < (1 << l); mask++) {
        c.a[l][mask] = a.a[l][mask] | b.a[l][mask];
      }
    }
    if (a.start.size() >= K) {
      c.start = a.start;
    } else {
      c.start = reduce(a.start + b.start);
    }
    if (b.last.size() >= K) {
      c.last = b.last;
    } else {
      c.last = reduce_reverse(a.last + b.last);
    }
    for (int i = 0; i < a.last.size(); i++) {
      for (int j = 0; j < b.start.size(); j++) {
        if (a.last.size() - i + j + 1 >= K) {
          continue;
        }
        int mask = 0, l = 0;
        for (int k = i; k < a.last.size(); k++) {
          mask |= (a.last[k] == '1') << l;
          l++;
        }
        for (int k = 0; k <= j; k++) {
          mask |= (b.start[k] == '1') << l;
          l++;
        }
        c.a[l][mask] = 1;
      }
    }
    return c;
  }
};
int n;
int main() {
  scanf("%d", &n);
  vector<Info> infos;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    Info info = Info(s);
    infos.push_back(info);
  }
  int m;
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--, b--;
    infos.push_back(infos[a] + infos[b]);
    printf("%d\n", infos.back().ans());
  }
  return 0;
}