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>
#pragma comment(linker, "/STACK:66777216")
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s1, s2;
  cin >> s1 >> s2;
  vector<int> bb1(s1.size() + 1);
  vector<int> bb2(s2.size() + 1);
  vector<int> aa1(s1.size() + 1);
  vector<int> aa2(s2.size() + 1);
  for (int i = 0; i < s1.size(); i++) {
    if (s1[i] != 'A')
      bb1[i + 1]++;
    else
      aa1[i + 1]++;
  }
  for (int i = 0; i < s2.size(); i++) {
    if (s2[i] != 'A')
      bb2[i + 1]++;
    else
      aa2[i + 1]++;
  }
  for (int i = 1; i < bb1.size(); i++) {
    bb1[i] += bb1[i - 1];
    if (s1[i - 1] == 'A') aa1[i] += aa1[i - 1];
  }
  for (int i = 1; i < bb2.size(); i++) {
    bb2[i] += bb2[i - 1];
    if (s2[i - 1] == 'A') aa2[i] += aa2[i - 1];
  }
  int q;
  cin >> q;
  vector<int> answer(q, 0);
  for (int i = 0; i < q; i++) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    int a1 = min(r1 - l1 + 1, aa1[r1]);
    int a2 = min(r2 - l2 + 1, aa2[r2]);
    int b1 = bb1[r1] - bb1[l1 - 1];
    int b2 = bb2[r2] - bb2[l2 - 1];
    if (a2 > a1 ||
        (r1 - l1 + 1 == a1 && a1 == a2 && r2 - l2 + 1 != r1 - l1 + 1)) {
      answer[i] = 0;
      continue;
    }
    if (a2 < a1 && (a1 - a2) % 3 != 0) {
      b1 += 2;
    }
    if (b1 > b2) {
      answer[i] = 0;
      continue;
    }
    if (b1 % 2 != b2 % 2) {
      answer[i] = 0;
      continue;
    }
    answer[i] = 1;
  }
  for (int i = 0; i < q; i++) {
    cout << answer[i];
  }
}
