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;
char st1[100011];
char st2[100011];
int s[100011][2];
int S[100011][2];
int n, m, i, j, k, l, Q, a, b, c, d, l1, r1, l2, r2;
int main() {
  scanf("%s", st1);
  scanf("%s", st2);
  n = strlen(st1);
  m = strlen(st2);
  for (i = 1; i <= n; i++) {
    s[i][0] = s[i - 1][0] + (st1[i - 1] != 'A');
    s[i][1] = (st1[i - 1] == 'A') ? s[i - 1][1] + 1 : 0;
  }
  for (i = 1; i <= m; i++) {
    S[i][0] = S[i - 1][0] + (st2[i - 1] != 'A');
    S[i][1] = (st2[i - 1] == 'A') ? S[i - 1][1] + 1 : 0;
  }
  for (scanf("%d", &Q); Q; Q--) {
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    a = s[r1][0] - s[l1 - 1][0];
    b = min(s[r1][1], r1 - l1 + 1);
    c = S[r2][0] - S[l2 - 1][0];
    d = min(S[r2][1], r2 - l2 + 1);
    if (a == c && b == d)
      printf("1");
    else if (a > c || b < d)
      printf("0");
    else if ((c - a) & 1)
      printf("0");
    else {
      if (b == d && a == 0)
        printf("0");
      else if (b > d && a + 2 <= c)
        printf("1");
      else if (!((b - d) % 3))
        printf("1");
      else
        printf("0");
    }
  }
  printf("\n");
}
