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 MAXN = 100;
const int MAXM = 100;
struct CADENA {
  string prefijo;
  string sufijo;
  bool la_tengo[10][(1 << 10)];
  void init(string s) {
    int n = s.size();
    if (n >= 10) {
      prefijo = s.substr(0, 10);
      sufijo = s.substr(n - 10);
    } else {
      prefijo = s;
      sufijo = s;
    }
    for (int k = 1; k < 10; k++) {
      for (int i = 0; i < (1 << k); i++) la_tengo[k][i] = false;
      if (k > n) continue;
      int num = 0;
      int f = 0;
      for (f = 0; f < k; f++) {
        num <<= 1;
        num |= (s[f] == '1');
      }
      la_tengo[k][num] = true;
      int mask_bit_apagar = (1 << (k - 1)) - 1;
      while (f < n) {
        num &= mask_bit_apagar;
        num <<= 1;
        num |= (s[f] == '1');
        la_tengo[k][num] = true;
        f++;
      }
    }
  }
  int obtenRespuesta() {
    int k;
    for (k = 9; k > 0; k--) {
      bool pasa = true;
      for (int i = 0; pasa && i < (1 << k); i++) {
        if (!la_tengo[k][i]) {
          pasa = false;
        }
      }
      if (pasa) break;
    }
    return k;
  }
  CADENA operator+(const CADENA& otro) const {
    CADENA ret;
    string nueva = sufijo + otro.prefijo;
    ret.init(nueva);
    ret.prefijo = prefijo;
    if (ret.prefijo.size() < 10) {
      int n = nueva.size();
      if (n < 10) {
        ret.prefijo = nueva;
      } else {
        ret.prefijo = nueva.substr(0, 10);
      }
    }
    ret.sufijo = otro.sufijo;
    if (ret.sufijo.size() < 10) {
      int n = nueva.size();
      if (n < 10) {
        ret.sufijo = nueva;
      } else {
        ret.sufijo = nueva.substr(n - 10);
      }
    }
    for (int k = 1; k < 10; k++) {
      for (int i = 0; i < (1 << k); i++) {
        ret.la_tengo[k][i] |= la_tengo[k][i] || otro.la_tengo[k][i];
      }
    }
    return ret;
  }
};
CADENA cadena[MAXN + MAXM + 2];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    cadena[i].init(s);
  }
  int m;
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    cadena[n + i] = cadena[a] + cadena[b];
    cout << cadena[n + i].obtenRespuesta() << "\n";
  }
  return 0;
}