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;
inline void read(int &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
const int N = 500000;
int n, m, qz1[N], qz2[N], lst1[N], lst2[N];
char s[N], t[N];
int main() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (register int i = 1; i <= n; i++) qz1[i] = qz1[i - 1] + (s[i] != 'A');
  for (register int i = 1; i <= m; i++) qz2[i] = qz2[i - 1] + (t[i] != 'A');
  for (register int i = 1; i <= n; i++)
    lst1[i] = (s[i] != 'A') ? i : lst1[i - 1];
  for (register int i = 1; i <= m; i++)
    lst2[i] = (t[i] != 'A') ? i : lst2[i - 1];
  int q;
  read(q);
  while (q--) {
    int a, b, c, d;
    read(a);
    read(b);
    read(c);
    read(d);
    int b1 = qz1[b] - qz1[a - 1], b2 = qz2[d] - qz2[c - 1],
        a1 = b - max(lst1[b], a - 1), a2 = d - max(lst2[d], c - 1);
    printf("%c", '0' + (b1 <= b2 && (b2 - b1) % 2 == 0 && a1 >= a2 &&
                        (b1 < b2 && (a1 - 1 >= a2 || a1 == a2 && b1 > 0) ||
                         b1 == b2 && (a1 - a2) % 3 == 0)));
  }
  return 0;
}
