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 st1[100010], st2[100010];
int a1[100010], a2[100010], b1[100010], b2[100010];
int n;
int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || '9' < ch) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int main() {
  scanf("%s%s", st1 + 1, st2 + 1);
  for (int i = 1; st1[i] != 0; i++) {
    if (st1[i] == 'A') a1[i] = a1[i - 1] + 1;
    b1[i] = b1[i - 1] + (st1[i] != 'A');
  }
  for (int i = 1; st2[i] != 0; i++) {
    if (st2[i] == 'A') a2[i] = a2[i - 1] + 1;
    b2[i] = b2[i - 1] + (st2[i] != 'A');
  }
  n = read();
  while (n--) {
    int l1 = read(), r1 = read(), l2 = read(), r2 = read(), ok = 1;
    int s1 = b1[r1] - b1[l1 - 1], s2 = b2[r2] - b2[l2 - 1];
    int x1 = min(r1 - l1 + 1, a1[r1]), x2 = min(r2 - l2 + 1, a2[r2]);
    if (s1 > s2 || (s1 + s2) % 2 == 1 || x1 < x2) ok = 0;
    if (s1 == s2 && x1 % 3 != x2 % 3 || s1 == 0 && x1 == x2 && s1 != s2) ok = 0;
    printf("%d", ok);
  }
}
