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;
int k, n, m, Q;
int s1[N], s2[N], f1[N], f2[N];
int l1, r1, l2, r2;
char s[N], t[N];
bool Solve(int l1, int r1, int l2, int r2) {
  int x = s1[r1] - s1[l1 - 1], y = s2[r2] - s2[l2 - 1];
  if (x > y || ((y - x) & 1)) return 0;
  int len1 = min(f1[r1], r1 - l1 + 1), len2 = min(f2[r2], r2 - l2 + 1);
  if (x == y) {
    if (len1 < len2) return 0;
    return !((len1 - len2) % 3);
  }
  if (!x) return len1 > len2;
  return len1 >= len2;
}
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%s", t + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; i++)
    s1[i] = s1[i - 1] + (s[i] != 'A'), f1[i] = s[i] == 'A' ? f1[i - 1] + 1 : 0;
  for (int i = 1; i <= m; i++)
    s2[i] = s2[i - 1] + (t[i] != 'A'), f2[i] = t[i] == 'A' ? f2[i - 1] + 1 : 0;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d", &l1);
    scanf("%d", &r1);
    scanf("%d", &l2);
    scanf("%d", &r2);
    putchar(Solve(l1, r1, l2, r2) ? '1' : '0');
  }
  return 0;
}
