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;
struct rua {
  char s[100001];
  int n, a[100001], b[100001], x, y;
  void _() {
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (s[i] == 'A')
        a[i] = ++cnt, b[i] = b[i - 1];
      else
        a[i] = cnt = 0, b[i] = b[i - 1] + 1;
  }
  void get(int l, int r) {
    x = b[r] - b[l - 1];
    if (!x)
      y = r - l + 1;
    else
      y = a[r];
  }
} s, t;
int Q, a, b, c, d;
int main() {
  scanf("%s", s.s + 1), s.n = strlen(s.s + 1);
  scanf("%s", t.s + 1), t.n = strlen(t.s + 1);
  s._(), t._();
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    s.get(a, b), t.get(c, d);
    if (!s.x && t.x)
      if (s.y > t.y)
        s.x += 2, s.y = t.y;
      else
        s.y--;
    if (s.y < t.y) {
      printf("0");
      continue;
    }
    if ((s.y - t.y) % 3) s.x += 2;
    if (s.x > t.x || (t.x - s.x) & 1) {
      printf("0");
      continue;
    }
    printf("1");
  }
  return printf("\n"), 0;
}
