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 = 1e5 + 5;
char s[N], p[N];
int n, m;
int a1[N], b1[N], a2[N], b2[N];
char ans[N];
bool solve(int b1, int a1, int b2, int a2) {
  if (b1 > b2 || a1 < a2 || (b2 - b1) % 2 || b2 && !b1 && !(a2 - a1)) return 0;
  if (b2 - b1) return 1;
  if ((a2 - a1) % 3 == 0) return 1;
  return 0;
}
int main() {
  scanf("%s%s", s + 1, p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (int i = 1; i <= n; i++) {
    b1[i] = b1[i - 1] + (s[i] != 'A');
    if (s[i] == 'A')
      a1[i] = a1[i - 1] + 1;
    else
      a1[i] = 0;
  }
  for (int i = 1; i <= m; i++) {
    b2[i] = b2[i - 1] + (p[i] != 'A');
    if (p[i] == 'A')
      a2[i] = a2[i - 1] + 1;
    else
      a2[i] = 0;
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    ans[i] = '0' + solve(b1[r1] - b1[l1 - 1], min(a1[r1], r1 - l1 + 1),
                         b2[r2] - b2[l2 - 1], min(a2[r2], r2 - l2 + 1));
  }
  ans[q] = '\0';
  printf("%s", ans);
}
