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 s[100010], t[100010];
int qzhsb[100010], qzhtb[100010], hztotsa[100010], hztotta[100010];
int n, m, q;
int main() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  scanf("%d", &q);
  for (int i = 1; i <= n; ++i) {
    qzhsb[i] = qzhsb[i - 1] + (s[i] != 'A');
    hztotsa[i] = (s[i] == 'A' ? hztotsa[i - 1] + 1 : 0);
  }
  for (int i = 1; i <= m; ++i) {
    qzhtb[i] = qzhtb[i - 1] + (t[i] != 'A');
    hztotta[i] = (t[i] == 'A' ? hztotta[i - 1] + 1 : 0);
  }
  for (int i = 1; i <= q; ++i) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int totsb = qzhsb[r1] - qzhsb[l1 - 1], tottb = qzhtb[r2] - qzhtb[l2 - 1],
        hzsa = min(hztotsa[r1], r1 - l1 + 1),
        hzta = min(hztotta[r2], r2 - l2 + 1);
    if ((hzta > hzsa) || ((tottb - totsb) & 1) ||
        (tottb - (((hzsa - hzta) % 3) > 0) * 2 < totsb) ||
        (totsb == 0 && tottb && hzsa == hzta))
      printf("0");
    else
      printf("1");
  }
}
