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[200 + 5], c[200 + 5], Head[200 + 5][200 + 5], End[200 + 5][200 + 5];
bitset<1024 + 5> Hash[200 + 5];
int n, m, len[200 + 5];
void Update(int x, int y, int k) {
  strcpy(c, End[x]), strcat(c, Head[y]);
  len[k] = len[x] + len[y];
  for (int i = 2; i <= min(len[k], 9); i++) {
    int t = (1 << i) - 1, cur = 0;
    for (int j = 0; j < i - 1; j++) cur = cur << 1 | (c[j] - '0');
    for (int j = i - 1; j < len[k]; j++) {
      cur = ((cur << 1) | (c[j] - '0')) & t;
      if (j >= len[x]) Hash[k][cur + (1 << i) - 2] = 1;
    }
  }
  if (len[x] < 9) {
    strcpy(Head[k], Head[x]);
    for (int j = len[x]; j < len[k]; j++) Head[k][j] = Head[y][j - len[x]];
  } else
    strcpy(Head[k], Head[x]);
  if (len[y] < 9) {
    for (int j = 0; j < 9 - len[y]; j++)
      End[k][j] = End[x][max(0, len[x] - 9 + len[y]) + j];
    strcat(End[k], End[y]);
  } else
    strcpy(End[k], End[y]);
  len[k] = min(len[k], 9);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s), len[i] = strlen(s);
    for (int j = 0; j < min(9, len[i]); j++)
      Head[i][j] = s[j], End[i][j] = s[max(0, len[i] - 9) + j];
    for (int j = 1; j <= 9; j++) {
      int t = (1 << j) - 1, cur = 0;
      for (int k = 0; k < j - 1; k++) cur = cur << 1 | (s[k] - '0');
      for (int k = j - 1; k < len[i]; k++) {
        cur = ((cur << 1) | (s[k] - '0')) & t;
        Hash[i][cur + (1 << j) - 2] = 1;
      }
    }
    len[i] = min(len[i], 9);
  }
  scanf("%d", &m);
  for (int i = n + 1; i <= n + m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    Hash[i] = Hash[x] | Hash[y];
    Update(x, y, i);
    for (int j = 0; j <= 1 << 10; j++)
      if (!Hash[i][j]) {
        printf("%d\n", (int)log2(j + 2) - 1);
        break;
      }
  }
  return 0;
}