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.
#include <bits/stdc++.h>
using namespace std;
char s[205][505];
int f[105];
int max(int a, int b) {
  if (a > b) return a;
  return b;
}
void work(int x) {
  int k;
  for (k = 1; k <= 10; k++) {
    for (int i = 0; i < (1 << k); i++) {
      char str[20] = {0};
      int t = i;
      for (int j = k - 1; j >= 0; j--) {
        str[j] = (t & 1) + '0';
        t >>= 1;
      }
      if (strstr(s[x], str) == NULL) {
        f[x] = k - 1;
        return;
      }
    }
  }
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    for (int i = 1; i <= n; i++) scanf("%s", s[i]);
    int sk[105] = {0};
    bool flag = 0;
    for (int i = 1; i <= n; i++) {
      for (int xx = 1; xx <= 10; xx++) {
        flag = 1;
        for (int j = 0; j < (1 << xx); j++) {
          char tmp[15] = {0};
          int tt = j;
          for (int ll = xx - 1; ll >= 0; ll--) {
            tmp[ll] = (tt & 1) + '0';
            tt = tt >> 1;
          }
          if (strstr(s[i], tmp) == NULL) flag = 0;
          if (flag == 0) break;
        }
        if (flag == 0) {
          sk[i] = xx - 1;
          break;
        }
      }
    }
    int m;
    scanf("%d", &m);
    int a[105] = {0}, b[105] = {0};
    for (int i = 1; i <= m; i++) {
      scanf("%d%d", &a[i], &b[i]);
      int j = 0;
      for (int x = max(0, strlen(s[a[i]]) - 200); s[a[i]][x] != 0; x++) {
        s[i + n][j] = s[a[i]][x];
        j++;
      }
      for (int x = 0; s[b[i]][x] != 0 && x < 200; x++) {
        s[i + n][j] = s[b[i]][x];
        j++;
      }
      work(i + n);
      memset(s[i + n], 0, sizeof(s[i + n]));
      j = 0;
      for (int x = 0; s[a[i]][x] != 0 && x < 200; x++) {
        s[i + n][j] = s[a[i]][x];
        j++;
      }
      for (int x = max(0, strlen(s[b[i]]) - 200); s[b[i]][x] != 0; x++) {
        s[i + n][j] = s[b[i]][x];
        j++;
      }
      f[i + n] = max(f[i + n], f[a[i]]);
      f[i + n] = max(f[i + n], f[b[i]]);
      printf("%d\n", f[i + n]);
    }
  }
  return 0;
}
