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;
int D[222][65536];
int n, L[222];
char p[222][222], tp[222];
char B[222][20], E[222][20];
void Make(int num, char *q, int L) {
  int i;
  for (i = 0; i < L && i < 16; i++) {
    B[num][i] = q[i];
    E[num][i] = q[L - 1 - i];
  }
}
int main() {
  int i, j, k, l, a, b;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%s", p[i]);
    for (j = 0; p[i][j]; j++)
      ;
    L[i] = j;
    for (j = 1; j <= 15; j++) {
      for (k = 0; k <= L[i] - j; k++) {
        int s = 0;
        for (l = k; l < k + j; l++) {
          s = s * 2 + p[i][l] - '0';
        }
        D[i][(1 << j) + s] = 1;
      }
    }
    Make(i, p[i], L[i]);
  }
  int Q;
  scanf("%d", &Q);
  for (i = n + 1; i <= n + Q; i++) {
    scanf("%d%d", &a, &b);
    for (j = 0; j < 65536; j++) {
      D[i][j] = D[a][j] | D[b][j];
    }
    int cnt = 0;
    for (j = min(16, L[a]) - 1; j >= 0; j--) tp[cnt++] = E[a][j];
    for (j = 0; j < L[b] && j < 16; j++) tp[cnt++] = B[b][j];
    for (j = 1; j <= 15; j++) {
      for (k = 0; k <= cnt - j; k++) {
        int s = 0;
        for (l = k; l < k + j; l++) {
          s = s * 2 + tp[l] - '0';
        }
        D[i][(1 << j) + s] = 1;
      }
    }
    int r = 0;
    for (j = 1; j <= 15; j++) {
      for (k = 0; k < (1 << j); k++) {
        if (!D[i][(1 << j) + k]) break;
      }
      if (k == (1 << j)) r = j;
    }
    printf("%d\n", r);
    cnt = 0;
    for (j = 0; j < L[a]; j++) tp[cnt++] = p[a][j];
    for (j = 0; j < L[b]; j++) tp[cnt++] = p[b][j];
    if (cnt < 60) {
      for (j = 0; j < cnt; j++) p[i][j] = tp[j];
      L[i] = cnt;
    } else {
      L[i] = 0;
      for (j = 0; j <= 20; j++) p[i][L[i]++] = tp[j];
      for (j = cnt - 20; j < cnt; j++) p[i][L[i]++] = tp[j];
    }
    Make(i, p[i], L[i]);
  }
}