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[1000000], b[1000000];
int c[1000000], d[1000000];
int e[1000000], f[1000000];
int main() {
  cin >> a + 1 >> b + 1;
  int n = strlen(a + 1);
  int m = strlen(b + 1);
  for (int i = 1; i <= n; i++) c[i] = c[i - 1] + (a[i] != 'A');
  for (int i = 1; i <= n; i++) e[i] = (e[i - 1] + 1) * (a[i] == 'A');
  for (int i = 1; i <= m; i++) d[i] = d[i - 1] + (b[i] != 'A');
  for (int i = 1; i <= m; i++) f[i] = (f[i - 1] + 1) * (b[i] == 'A');
  int q, l1, r1, l2, r2, E, F, C, D;
  cin >> q;
  while (q--) {
    cin >> l1 >> r1 >> l2 >> r2;
    E = e[r1];
    if (E > r1 - l1 + 1) E = r1 - l1 + 1;
    F = f[r2];
    if (F > r2 - l2 + 1) F = r2 - l2 + 1;
    C = c[r1] - c[l1 - 1];
    D = d[r2] - d[l2 - 1];
    if (D && !C) {
      E--;
      C = 2;
    } else if ((E - F) % 3)
      C += 2;
    cout << (C <= D && ((D - C) % 2 == 0) && E >= F);
  }
  return 0;
}
