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[100005];
int A[100005], B[100005], C[100005], D[100005], n, t1, t2;
int main() {
  scanf("%s", s + 1);
  for (int i = 1; s[i]; ++i) {
    A[i] = (s[i] != 'A') + A[i - 1];
    C[i] = (s[i] == 'A' ? 1 + C[i - 1] : 0);
  }
  scanf("%s", s + 1);
  for (int i = 1; s[i]; ++i) {
    B[i] = (s[i] != 'A') + B[i - 1];
    D[i] = (s[i] == 'A' ? 1 + D[i - 1] : 0);
  }
  scanf("%d", &n);
  s[n] = 0;
  for (int i = 0, a, b, c, d; i < n; ++i) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    t1 = min(C[b], b - a + 1) - min(D[d], d - c + 1);
    t2 = A[b] - A[a - 1] - B[d] + B[c - 1];
    if (!t2 && t1 > 0 && t1 % 3) t2 = 1;
    if (!t1 && !(A[b] - A[a - 1]) && B[d] - B[c - 1]) t2 = 1;
    if (!D[d] && t1) t2 += (t1 % 3 ? 2 : 0), t1 = 0;
    if (t1 > 0) t1 = 0, t2 += (t1 % 3 ? 2 : 0);
    s[i] = '1' - (t1 < 0 || t1 % 3 || t2 & 1 || t2 > 0);
  }
  return !printf("%s\n", s);
}
