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>
using namespace std;
int n, m, q, l, r, b0, a0, b1, a1, sum[100005 << 1], lst[100005 << 1];
char s[100005 << 1];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%s%d", s + n + 1, &q);
  m = strlen(s + 1);
  for (int i = 1; i <= m; i++)
    sum[i] = sum[i - 1] + (s[i] != 'A'),
    lst[i] = (s[i] == 'A' ? lst[i - 1] + 1 : 0);
  while (q--) {
    scanf("%d%d", &l, &r), a0 = min(lst[r], r - l + 1),
                           b0 = sum[r] - sum[l - 1];
    scanf("%d%d", &l, &r), l += n, r += n, a1 = min(r - l + 1, lst[r]),
                                           b1 = sum[r] - sum[l - 1];
    if (b0 > b1 || ((b1 - b0) & 1))
      putchar('0');
    else if (b0 == b1)
      putchar(a0 - a1 >= 0 && (a0 - a1) % 3 == 0 ? '1' : '0');
    else if (b0)
      putchar(a1 > a0 ? '0' : '1');
    else
      putchar(a0 > a1 ? '1' : '0');
  }
  return 0;
}
