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 n, m;
char s[100];
int l[202], v1[202][15], v2[202][15], v[30];
bitset<32768> b[202][16];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s);
    int len = strlen(s);
    l[i] = min(15, len);
    for (int j = 0; j < l[i]; ++j) {
      v1[i][j] = s[j] - '0';
      v2[i][j] = s[len - j - 1] - '0';
    }
    for (int j = 0; j < len; ++j) {
      int st = 0;
      for (int k = 1; k <= 15 && j + k - 1 < len; ++k) {
        st |= (s[j + k - 1] - '0') << (k - 1);
        b[i][k][st] = 1;
      }
    }
  }
  scanf("%d", &m);
  for (int i = 1, x, y; i <= m; ++i) {
    scanf("%d%d", &x, &y);
    l[++n] = min(15, l[x] + l[y]);
    for (int j = 0; j < l[n]; ++j) {
      v1[n][j] = j < l[x] ? v1[x][j] : v1[y][j - l[x]];
      v2[n][j] = j < l[y] ? v2[y][j] : v2[x][j - l[y]];
    }
    for (int j = 1; j <= 15; ++j) b[n][j] = b[x][j] | b[y][j];
    int len = 0;
    for (int j = l[x] - 1; j >= 0; --j) v[len++] = v2[x][j];
    for (int j = 0; j < l[y]; ++j) v[len++] = v1[y][j];
    for (int j = 0; j < len; ++j) {
      int st = 0;
      for (int k = 1; k <= 15 && j + k - 1 < len; ++k) {
        st |= v[j + k - 1] << (k - 1);
        b[n][k][st] = 1;
      }
    }
    int ans = 0;
    for (int j = 15; j; --j) {
      int ok = 1;
      for (int k = 0; k < (1 << j); ++k)
        if (!b[n][j][k]) {
          ok = 0;
          break;
        }
      if (ok) {
        ans = j;
        break;
      }
    }
    printf("%d\n", ans);
  }
}