Alice has a string consisting of characters 'A', 'B' and 'C'. Bob can use the following transitions on any substring of our string in any order any number of times: 

  * A <image> BC
  * B <image> AC
  * C <image> AB
  * AAA <image> empty string 



Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.

Input

The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters 'A', 'B' and 'C'.

The third line contains the number of queries Q (1 ≤ Q ≤ 105).

The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?

Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.

It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.

Output

Print a string of Q characters, where the i-th character is '1' if the answer to the i-th query is positive, and '0' otherwise.

Example

Input

AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3


Output

10011

Note

In the first query we can achieve the result, for instance, by using transitions <image>.

The third query asks for changing AAB to A — but in this case we are not able to get rid of the character 'B'.
#include <bits/stdc++.h>
const int maxline = 100001;
int main() {
  char str;
  int n, m, q, i, j, k;
  int s[maxline], t[maxline], a, b, c, d;
  int f[2][maxline], g[2][maxline];
  str = getchar();
  n = 0;
  int sum = 0;
  f[0][0] = 0;
  g[0][0] = 0;
  while (str == 'A' || str == 'B' || str == 'C') {
    n++;
    if (str == 'A') {
      s[n] = 0;
      sum++;
      f[0][n] = f[0][n - 1];
      g[0][n] = sum;
    } else {
      s[n] = 1;
      sum = 0;
      f[0][n] = f[0][n - 1] + 1;
      g[0][n] = sum;
    }
    str = getchar();
  }
  while (str != 'A' && str != 'B' && str != 'C') str = getchar();
  m = 0;
  sum = 0;
  f[1][0] = 0;
  g[1][0] = 0;
  while (str == 'A' || str == 'B' || str == 'C') {
    m++;
    if (str == 'A') {
      t[m] = 0;
      sum++;
      f[1][m] = f[1][m - 1];
      g[1][m] = sum;
    } else {
      t[m] = 1;
      sum = 0;
      f[1][m] = f[1][m - 1] + 1;
      g[1][m] = sum;
    }
    str = getchar();
  }
  scanf("%d", &q);
  int answer[maxline];
  for (i = 1; i <= q; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    answer[i] = 0;
    int e = (f[1][d] - f[1][c - 1]) - (f[0][b] - f[0][a - 1]);
    int es, et, ok1 = 1, ok2 = 1;
    if (f[0][b] - f[0][a - 1] > 0)
      es = g[0][b];
    else {
      es = b - a + 1;
      ok1 = 0;
    }
    if (f[1][d] - f[1][c - 1] > 0)
      et = g[1][d];
    else {
      et = d - c + 1;
      ok2 = 0;
    }
    if (!(ok1 == 0 && ok2 == 1) || es - et > 0)
      if (e >= 0 && e % 2 == 0 && es - et >= 0) {
        if (e > 0)
          answer[i] = 1;
        else if ((es - et) % 3 == 0)
          answer[i] = 1;
      }
  }
  for (i = 1; i <= q; i++) printf("%d", answer[i]);
  printf("\n");
}
