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 str[2][110000];
int n[2], sum[2][110000], g[2][110000], num_q, a, b, c, d;
int main() {
  scanf("%s%s", str[0], str[1]);
  int i, j, s, p, q;
  n[0] = strlen(str[0]);
  n[1] = strlen(str[1]);
  for (i = 0; i < 2; i++) {
    sum[i][0] = g[i][0] = 0;
    for (j = 0; j < n[i]; j++) {
      sum[i][j + 1] = sum[i][j];
      if (str[i][j] == 'A') {
        sum[i][j + 1]++;
        g[i][j + 1] = g[i][j] + 1;
      } else
        g[i][j + 1] = 0;
    }
  }
  scanf("%d", &num_q);
  while (num_q--) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    bool ans = 0;
    if (min(g[0][b], b - a + 1) >= min(g[1][d], d - c + 1)) {
      int ld = (b - a) - (d - c) - (sum[0][b] - sum[0][a - 1]);
      if ((min(g[0][b], b - a + 1) % 3 != min(g[1][d], d - c + 1) % 3)) ld += 2;
      ld += sum[1][d] - sum[1][c - 1];
      if (ld <= 0 && ld % 2 == 0) ans = 1;
      if (min(g[0][b], b - a + 1) == b - a + 1 &&
          min(g[1][d], d - c + 1) == b - a + 1 &&
          min(g[1][d], d - c + 1) != d - c + 1)
        ans = 0;
    }
    printf("%d", ans);
  }
  printf("\n");
  return 0;
}
