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>
int first_not_a_cnt[100001], second_not_a_cnt[100001], first_last_not_a[100001],
    second_last_not_a[100001];
int max(int a, int b) { return a < b ? b : a; }
int main() {
  std::string first_input_string, second_input_string;
  std::cin >> first_input_string >> second_input_string;
  first_not_a_cnt[0] = second_not_a_cnt[0] = 0;
  for (int i = 1; i <= first_input_string.size(); ++i) {
    first_not_a_cnt[i] =
        first_not_a_cnt[i - 1] + (first_input_string[i - 1] == 'A' ? 0 : 1);
  }
  for (int i = 1; i <= second_input_string.size(); ++i) {
    second_not_a_cnt[i] =
        second_not_a_cnt[i - 1] + (second_input_string[i - 1] == 'A' ? 0 : 1);
  }
  first_last_not_a[0] = second_last_not_a[0] = 0;
  for (int i = 1; i <= first_input_string.size(); ++i) {
    if (first_input_string[i - 1] != 'A') {
      first_last_not_a[i] = i;
    } else {
      first_last_not_a[i] = first_last_not_a[i - 1];
    }
  }
  for (int i = 1; i <= second_input_string.size(); ++i) {
    if (second_input_string[i - 1] != 'A') {
      second_last_not_a[i] = i;
    } else {
      second_last_not_a[i] = second_last_not_a[i - 1];
    }
  }
  int num_queries;
  std::cin >> num_queries;
  int first_begin, first_end, second_begin, second_end;
  for (int q0 = 0; q0 < num_queries; ++q0) {
    std::cin >> first_begin >> first_end >> second_begin >> second_end;
    int first_not_a_num =
            first_not_a_cnt[first_end] - first_not_a_cnt[first_begin - 1],
        second_not_a_num =
            second_not_a_cnt[second_end] - second_not_a_cnt[second_begin - 1],
        first_ending_a_cnt =
            first_end - max(first_begin, first_last_not_a[first_end]),
        second_ending_a_cnt =
            second_end - max(second_begin, second_last_not_a[second_end]);
    if (second_not_a_num < first_not_a_num ||
        ((second_not_a_num - first_not_a_num) & 1) ||
        (first_ending_a_cnt < second_ending_a_cnt)) {
      std::cout << 0;
      continue;
    }
    if (first_not_a_num != second_not_a_num) {
      std::cout << 1;
      continue;
    }
    if (second_ending_a_cnt <= first_ending_a_cnt &&
        ((first_ending_a_cnt - second_ending_a_cnt) % 3 == 0)) {
      std::cout << 1;
    } else {
      std::cout << 0;
    }
  }
  std::cout << std::endl;
  return 0;
}
