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;
char s[105][105];
char pre[205][10], nxt[205][10];
set<int> S[205][10];
int main() {
  int n = 0, q, a, b;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s[i]);
    int m = strlen(s[i]);
    for (int j = 0; j < m; ++j)
      for (int k = j; k < j + 9 && k < m; ++k) {
        int len = k - j + 1, t = 0;
        for (int l = j; l <= k; ++l) t += ((s[i][l] - '0') << (l - j));
        S[i][len].insert(t);
      }
    for (int j = 0; j < 8 && j < m; ++j) pre[i][j] = s[i][j];
    int l = 0;
    for (int j = max(0, m - 8); j < m; ++j) nxt[i][l++] = s[i][j];
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &a, &b);
    n++;
    for (int i = 1; i <= 9; ++i)
      for (set<int>::iterator it = S[a][i].begin(); it != S[a][i].end(); ++it)
        S[n][i].insert(*it);
    for (int i = 1; i <= 9; ++i)
      for (set<int>::iterator it = S[b][i].begin(); it != S[b][i].end(); ++it)
        S[n][i].insert(*it);
    int pa = strlen(pre[a]);
    int na = strlen(nxt[a]);
    int pb = strlen(pre[b]);
    int nb = strlen(nxt[b]);
    int m = 0;
    if (nb < 8)
      for (int i = max(0, na - (8 - nb)); i < na; ++i) nxt[n][m++] = nxt[a][i];
    for (int i = max(0, nb - 8); i < nb; ++i) nxt[n][m++] = nxt[b][i];
    m = 0;
    for (int i = 0; i < 8 && i < pa; ++i) pre[n][m++] = pre[a][i];
    if (m < 8)
      for (int i = 0; i < pb && m < 8; ++i) pre[n][m++] = pre[b][i];
    for (int i = max(0, na - 8); i < na; ++i) {
      int t = 0;
      for (int k = i; k < na; ++k) t += ((nxt[a][k] - '0') << (k - i));
      for (int j = 0; j < pb && j < 8; ++j) {
        int l = j + 1 + na - i;
        if (l > 9) continue;
        int t1 = t;
        for (int k = 0; k <= j; ++k) t1 += ((pre[b][k] - '0') << (na - i + k));
        S[n][l].insert(t1);
      }
    }
    for (int i = 9; i >= 0; --i)
      if (!i || (S[n][i].size() == (1 << i))) {
        printf("%d\n", i);
        break;
      }
  }
  return 0;
}