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 txt[2][200000];
int A[2][100000], B[2][100000];
int check(int bs, int bt, int as, int at) {
  if ((bs - bt) % 2) return 0;
  if (bs > bt) return 0;
  if (bs > 0) {
    if (as == at) return 1;
    if (as < at) return 0;
    if (bs < bt) return 1;
    return (as - at) % 3 == 0;
  }
  if (bt == 0) return as >= at && (as - at) % 3 == 0;
  return as > at;
}
int main() {
  int i, j, q, u, v, w, x, AS, AT, BS, BT;
  scanf("%s%s", txt[0], txt[1]);
  for (i = 0; i <= 1; i++) {
    for (j = 0; txt[i][j]; j++) {
      if (txt[i][j] != 'A') {
        B[i][j] = 1 + (j == 0 ? 0 : B[i][j - 1]);
        A[i][j] = 0;
      } else {
        A[i][j] = 1 + (j == 0 ? 0 : A[i][j - 1]);
        B[i][j] = (j == 0 ? 0 : B[i][j - 1]);
      }
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d%d", &u, &v, &w, &x);
    u--, v--, w--, x--;
    BS = B[0][v] - (u == 0 ? 0 : B[0][u - 1]);
    BT = B[1][x] - (w == 0 ? 0 : B[1][w - 1]);
    AS = min(A[0][v], v - u + 1);
    AT = min(A[1][x], x - w + 1);
    printf("%d", check(BS, BT, AS, AT));
  }
  printf("\n");
  return 0;
}
