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;
const int N = 100010;
struct SS {
  char s[N];
  int sum[N], l, a[N];
  void init() {
    l = strlen(s + 1);
    for (int i = 0; i <= l; ++i) {
      if (s[i] == 'A')
        sum[i] = sum[i - 1], a[i] = a[i - 1] + 1;
      else
        sum[i] = sum[i - 1] + 1, a[i] = 0;
    }
  }
} A, B;
int T;
int main() {
  scanf("%s%s", A.s + 1, B.s + 1);
  A.init();
  B.init();
  scanf("%d", &T);
  int lA, rA, lB, rB, t1, t2;
  while (T--) {
    scanf("%d%d%d%d", &lA, &rA, &lB, &rB);
    lA--;
    lB--;
    t1 = (B.sum[rB] - B.sum[lB]) - (A.sum[rA] - A.sum[lA]);
    t2 = min(rA - lA, A.a[rA]) - min(rB - lB, B.a[rB]);
    if (t2 < 0 || t1 < 0 || t1 & 1 || (t2 % 3 != 0 && t1 == 0) ||
        (t2 == 0 && rA - lA <= A.a[rA] && t1 != 0)) {
      printf("0");
    } else
      printf("1");
  }
}
