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 A[200000 + 5];
int B[200000 + 5];
string str1, str2;
void pre() {
  int size1 = str1.size();
  for (int i = 0; i < size1; i++) {
    if (str1[i] != 'A') A[i + 1]++;
    A[i + 2] = A[i + 1];
  }
  int size2 = str2.size();
  for (int i = 0; i < size2; i++) {
    if (str2[i] != 'A') B[i + 1]++;
    B[i + 2] = B[i + 1];
  }
}
int main() {
  string ans;
  cin >> str1 >> str2;
  int n;
  cin >> n;
  pre();
  int a, b, c, d;
  int num1, num2;
  int X, Y, sum1, sum2;
  for (int i = 0; i < n; i++) {
    cin >> a >> b >> c >> d;
    num1 = A[b] - A[a - 1];
    num2 = B[d] - B[c - 1];
    X = b - a, Y = d - c;
    if (num1 + num2 == 0) {
      if (X < Y)
        ans += '0';
      else if (((X) - (Y)) % 3 == 0)
        ans += '1';
      else
        ans += '0';
    } else if (abs(num1 - num2) % 2 == 0 && num1 <= num2) {
      sum1 = 0;
      sum2 = 0;
      int dis = (num2 - num1) >> 1;
      for (int i = b - 1; i >= a - 1; i--) {
        if (str1[i] == 'A')
          sum1++;
        else
          break;
      }
      for (int i = d - 1; i >= c - 1; i--) {
        if (str2[i] == 'A')
          sum2++;
        else
          break;
      }
      if (num1 == 0) {
        if (sum1 > sum2)
          ans += '1';
        else
          ans += '0';
      } else if (sum2 > sum1)
        ans += '0';
      else if (num1 == num2) {
        if ((sum1 - sum2) % 3 == 0)
          ans += '1';
        else
          ans += '0';
      } else if (dis > 0)
        ans += '1';
      else
        ans += '0';
    } else
      ans += '0';
  }
  cout << ans;
}
