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 maxk = 16;
struct Node {
  string st;
  string ed;
  set<string> table[maxk + 1];
};
Node Union(Node& a, Node& b) {
  Node ret;
  for (int i = 1; i <= maxk; i++) {
    for (set<string>::iterator p = a.table[i].begin(); p != a.table[i].end();
         ++p) {
      ret.table[i].insert(*p);
    }
    for (set<string>::iterator p = b.table[i].begin(); p != b.table[i].end();
         ++p) {
      ret.table[i].insert(*p);
    }
  }
  string s = a.ed + b.st;
  int l = s.length();
  for (int len = 1; len <= maxk; len++) {
    for (int i = 0; i < l - len + 1; i++) {
      ret.table[len].insert(s.substr(i, len));
    }
  }
  if (l <= maxk) {
    ret.st = s;
    ret.ed = s;
  } else {
    ret.st = s.substr(0, maxk);
    ret.ed = s.substr(l - maxk);
  }
  if (a.st.length() == maxk) {
    ret.st = a.st;
  }
  if (b.ed.length() == maxk) {
    ret.ed = b.ed;
  }
  return ret;
}
int check(Node& a) {
  for (int k = maxk; k > 0; k--) {
    if (a.table[k].size() == 1 << k) {
      return k;
    }
  }
  return 0;
}
vector<Node> a;
int n;
int main() {
  cin >> n;
  a.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    int l = s.length();
    for (int len = 1; len <= maxk; len++) {
      for (int j = 0; j <= l - len; j++) {
        a[i].table[len].insert(s.substr(j, len));
      }
    }
    if (l <= maxk) {
      a[i].st = s;
      a[i].ed = s;
    } else {
      a[i].st = s.substr(0, maxk);
      a[i].ed = s.substr(l - maxk, maxk);
    }
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    a.push_back(Union(a[x], a[y]));
    cout << check(a.back()) << endl;
  }
  return 0;
}