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;
string S, T;
int S_B[100000 + 100], S_A[100000 + 100];
int T_B[100000 + 100], T_A[100000 + 100];
void calc(string str, int *B, int *A) {
  memset(B, 0, sizeof(B));
  memset(A, 0, sizeof(A));
  for (int i = 0; i < str.length(); i++) {
    if (str[i] != 'A') {
      B[i + 1]++;
    } else {
      A[i + 1] = A[i] + 1;
    }
    B[i + 1] += B[i];
  }
}
void init() {
  calc(S, S_B, S_A);
  calc(T, T_B, T_A);
}
void solve() {
  init();
  int q;
  cin >> q;
  int a, b, c, d;
  string res;
  while (q--) {
    scanf("%d %d %d %d\n", &a, &b, &c, &d);
    int B1 = S_B[b] - S_B[a - 1];
    int A1 = min(S_A[b], b - a + 1);
    int B2 = T_B[d] - T_B[c - 1];
    int A2 = min(T_A[d], d - c + 1);
    if (B1 == B2 && A1 >= A2 && (A1 - A2) % 3 == 0) {
      res += '1';
    } else if (B1 < B2 && (B2 - B1) % 2 == 0) {
      if ((B1 > 0 && A1 >= A2) || (B1 == 0 && A1 > A2)) {
        res += '1';
      } else {
        res += '0';
      }
    } else {
      res += '0';
    }
  }
  cout << res << endl;
}
int main() {
  cin >> S >> T;
  solve();
}
