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;
int val[4][100100];
void solvre(int a, int b, int c, int d) {
  int x = val[0][b] - val[0][a - 1];
  int y = val[2][d] - val[2][c - 1];
  int A = val[1][b] - val[1][a - 1];
  int B = val[3][d] - val[3][c - 1];
  if (val[1][a - 1] > 0 && val[1][b] - val[1][a - 1] == b - a + 1) {
    A = b - a + 1;
  } else
    A = val[1][b];
  if (val[3][c - 1] > 0 && val[3][d] - val[3][c - 1] == d - c + 1) {
    B = d - c + 1;
  } else
    B = val[3][d];
  if (x == 0 && y > 0 && A > 0) {
    A--;
  }
  if ((y - x) % 2 != 0 || y < x) {
    cout << 0;
    return;
  }
  int tirar = (y - x) / 2;
  if (tirar > 0) tirar = 19;
  if (B <= A && (A - B) % 3 <= tirar)
    cout << 1;
  else
    cout << 0;
}
int main() {
  string s, t;
  cin >> s >> t;
  int a, b, c, d;
  int n;
  cin >> n;
  val[0][0] = val[1][0] = val[2][0] = val[3][0] = 0;
  for (int i = 0; i < s.length(); i++) {
    if (s[i] != 'A') {
      val[1][i + 1] = 0;
      val[0][i + 1] = val[0][i] + 1;
    } else {
      val[1][i + 1] = val[1][i] + 1;
      val[0][i + 1] = val[0][i];
    }
  }
  for (int i = 0; i < t.length(); i++) {
    if (t[i] != 'A') {
      val[3][i + 1] = 0;
      val[2][i + 1] = val[2][i] + 1;
    } else {
      val[3][i + 1] = val[3][i] + 1;
      val[2][i + 1] = val[2][i];
    }
  }
  while (n--) {
    cin >> a >> b >> c >> d;
    solvre(a, b, c, d);
  }
  cout << endl;
}
