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;
const int mod = (1e9) + 7;
const double eps = 1e-9;
const int siz = 1e5 + 5, siz2 = 1e5 + 5;
char st[2][siz];
int q, bc[2][siz], lbc[2][siz], sz[2];
int main() {
  scanf("%s%s%d", st[0] + 1, st[1] + 1, &q);
  sz[0] = strlen(st[0] + 1);
  sz[1] = strlen(st[1] + 1);
  for (int i = 0; i < 2; i++) {
    for (int j = 1; j <= sz[i]; j++) {
      bc[i][j] = bc[i][j - 1];
      lbc[i][j] = lbc[i][j - 1];
      if (st[i][j] != 'A') {
        bc[i][j]++;
        lbc[i][j] = j;
      }
    }
  }
  vector<int> ans;
  while (q--) {
    int ai, bi, ci, di;
    scanf("%d%d%d%d", &ai, &bi, &ci, &di);
    int bc0 = bc[0][bi] - bc[0][ai - 1], bc1 = bc[1][di] - bc[1][ci - 1];
    int la0 = bi - max(ai - 1, lbc[0][bi]), la1 = di - max(ci - 1, lbc[1][di]);
    if (bc0 == 0 && bc1 != 0) {
      la0 = min(la0 - 1, la1);
      bc0 += 2;
    }
    if (bc1 >= bc0 && (bc1 - bc0) % 2 == 0 && la0 >= la1 &&
        ((la0 - la1) % 3 == 0 || bc1 != bc0))
      ans.push_back(1);
    else
      ans.push_back(0);
  }
  for (auto v : ans) printf("%d", v);
  printf("\n");
  return 0;
}
