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;
char S[100005], T[100005];
int q;
int Ssum[100005], Tsum[100005];
int Ssuffix[100005], Tsuffix[100005];
int main() {
  scanf("%s", S + 1);
  scanf("%s", T + 1);
  for (int i = 1; S[i]; i++) {
    Ssum[i] = Ssum[i - 1];
    if (S[i] != 'A') {
      Ssum[i]++;
      Ssuffix[i] = 0;
    } else {
      Ssuffix[i] = Ssuffix[i - 1] + 1;
    }
  }
  for (int i = 1; T[i]; i++) {
    Tsum[i] = Tsum[i - 1];
    if (T[i] != 'A') {
      Tsum[i]++;
      Tsuffix[i] = 0;
    } else {
      Tsuffix[i] = Tsuffix[i - 1] + 1;
    }
  }
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int l1, r1, l2, r2, a, b, x, y;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    a = Ssuffix[r1];
    a = min(a, r1 - l1 + 1);
    b = Tsuffix[r2];
    b = min(b, r2 - l2 + 1);
    x = Ssum[r1] - Ssum[l1 - 1];
    y = Tsum[r2] - Tsum[l2 - 1];
    if (a < b) {
      printf("0");
    } else {
      if (a == b && x == 0 && y != 0) {
        printf("0");
        continue;
      }
      if ((a - b) % 3 != 0) x += 2;
      printf("%d", y >= x && (y - x) % 2 == 0 ? 1 : 0);
    }
  }
  return 0;
}
