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 a[210000], b[210000];
int qa[210000], qb[210000], qc[210000], qd[210000];
int main() {
  scanf("%s", a + 1);
  scanf("%s", b + 1);
  int lena = strlen(a + 1), lenb = strlen(b + 1);
  qa[0] = 0;
  for (int i = 1; i <= lena; i++) qa[i] = qa[i - 1] + (a[i] != 'A');
  qb[0] = 0;
  for (int i = 1; i <= lenb; i++) qb[i] = qb[i - 1] + (b[i] != 'A');
  qc[0] = 0;
  for (int i = 1; i <= lena; i++) qc[i] = a[i] != 'A' ? 0 : qc[i - 1] + 1;
  qd[0] = 0;
  for (int i = 1; i <= lenb; i++) qd[i] = b[i] != 'A' ? 0 : qd[i - 1] + 1;
  int q;
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int temp = qa[r1] - qa[l1 - 1], temp2 = qb[r2] - qb[l2 - 1], temp1, temp3;
    if (r1 - qc[r1] < l1)
      temp1 = r1 - l1 + 1;
    else
      temp1 = qc[r1];
    if (r2 - qd[r2] < l2)
      temp3 = r2 - l2 + 1;
    else
      temp3 = qd[r2];
    if (temp2 < temp)
      printf("0");
    else if (temp2 % 2 != temp % 2)
      printf("0");
    else if (temp1 < temp3)
      printf("0");
    else if (temp1 == temp3 && temp == 0 && temp2 > 0)
      printf("0");
    else if (temp == temp2 && temp1 % 3 != temp3 % 3)
      printf("0");
    else
      printf("1");
  }
  printf("\n");
  return 0;
}
