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;
char s[100005], t[100005];
int o[2][100005], b[2][100005], p[2][100005];
int main() {
  int i, j, k, l, tmp, ii;
  scanf("%s", s);
  scanf("%s", t);
  for (i = 1; s[i - 1] != '\0'; i++) {
    if (s[i - 1] == 'A') {
      o[0][i] = o[0][i - 1] + 1;
      b[0][i] = b[0][i - 1];
      p[0][i] = p[0][i - 1];
    } else {
      p[0][i] = i;
      o[0][i] = o[0][i - 1];
      b[0][i] = b[0][i - 1] + 1;
    }
  }
  for (i = 1; t[i - 1] != '\0'; i++) {
    if (t[i - 1] == 'A') {
      o[1][i] = o[1][i - 1] + 1;
      b[1][i] = b[1][i - 1];
      p[1][i] = p[1][i - 1];
    } else {
      p[1][i] = i;
      o[1][i] = o[1][i - 1];
      b[1][i] = b[1][i - 1] + 1;
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%d %d %d", &i, &j, &k);
    scanf("%d", &l);
    if (b[0][j] - b[0][i - 1] > b[1][l] - b[1][k - 1] ||
        (b[1][l] - b[1][k - 1] - (b[0][j] - b[0][i - 1])) % 2 == 1) {
      printf("0");
      continue;
    }
    tmp = b[1][l] - b[1][k - 1] - (b[0][j] - b[0][i - 1]);
    ii = i;
    i = max(i - 1, p[0][j]);
    k = max(k - 1, p[1][l]);
    if (o[0][j] - o[0][i] < o[1][l] - o[1][k]) {
      printf("0");
      continue;
    } else if (tmp == 0 && (o[0][j] - o[0][i] - (o[1][l] - o[1][k])) % 3 != 0) {
      printf("0");
      continue;
    } else if (tmp > 0 && o[0][j] - o[0][i] == o[1][l] - o[1][k] &&
               b[0][j] - b[0][ii - 1] == 0) {
      printf("0");
      continue;
    }
    printf("1");
  }
  return 0;
}
