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[100010], b[100010];
int f0[100010], f1[100010], f2[100010], f3[100010];
int main() {
  int n, m, q, i, j, l1, l2, r1, r2, a0, a1, a2, a3;
  scanf("%s%s", a + 1, b + 1);
  n = strlen(a + 1), m = strlen(b + 1);
  for (i = 1; i <= n; i++)
    f0[i] = f0[i - 1] + (a[i] != 'A'), f2[i] = a[i] != 'A' ? 0 : f2[i - 1] + 1;
  for (i = 1; i <= m; i++)
    f1[i] = f1[i - 1] + (b[i] != 'A'), f3[i] = b[i] != 'A' ? 0 : f3[i - 1] + 1;
  for (scanf("%d", &q); q > 0; q--) {
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    a0 = f0[r1] - f0[l1 - 1], a2 = f1[r2] - f1[l2 - 1];
    if (r1 - f2[r1] < l1)
      a1 = r1 - l1 + 1;
    else
      a1 = f2[r1];
    if (r2 - f3[r2] < l2)
      a3 = r2 - l2 + 1;
    else
      a3 = f3[r2];
    if (a2 < a0)
      putchar('0');
    else if ((a2 - a0) % 2 != 0)
      putchar('0');
    else if (a1 < a3)
      putchar('0');
    else if (a1 == a3 && a0 == 0 && a2 > 0)
      putchar('0');
    else if (a0 == a2 && (a1 - a3) % 3 != 0)
      putchar('0');
    else
      putchar('1');
  }
  putchar('\n');
  return 0;
}
