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;
int fs[100001], ft[100001], gs[100001], gt[100001];
int main() {
  char c1;
  int n = 0, m = 0, q, a, b, c, d, ans;
  scanf("%c", &c1);
  while (c1 >= 'A' && c1 <= 'C') {
    n++;
    if (c1 != 'A')
      fs[n] = fs[n - 1] + 1;
    else
      fs[n] = fs[n - 1];
    if (c1 == 'A') gs[n] = gs[n - 1] + 1;
    scanf("%c", &c1);
  }
  scanf("%c", &c1);
  while (c1 >= 'A' && c1 <= 'C') {
    m++;
    if (c1 != 'A')
      ft[m] = ft[m - 1] + 1;
    else
      ft[m] = ft[m - 1];
    if (c1 == 'A') gt[m] = gt[m - 1] + 1;
    scanf("%c", &c1);
  }
  scanf("%d", &q);
  for (int k = 1; k <= q; k++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    ans = 1;
    int a1 = fs[b] - fs[a - 1];
    int b1 = min(b - a + 1, gs[b]);
    int c1 = ft[d] - ft[c - 1];
    int d1 = min(d - c + 1, gt[d]);
    if (a1 > c1) ans = 0;
    if ((c1 - a1) % 2 == 1) ans = 0;
    if (b1 < d1) ans = 0;
    if ((a1 == 0) && (c1 > 0) && (b1 == d1)) ans = 0;
    if ((a1 == c1) && ((b1 % 3) != (d1 % 3))) ans = 0;
    printf("%d", ans);
  }
}
