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 a[100005], b[100005];
int c[2][100005], h[2][100005];
int main() {
  scanf("%s%s", a + 1, b + 1);
  int l1 = strlen(a + 1), l2 = strlen(b + 1);
  for (int i = 1; i <= l1; i++) {
    c[0][i] = c[0][i - 1];
    h[0][i] = h[0][i - 1];
    if (a[i] != 'A')
      c[0][i]++, h[0][i] = 0;
    else
      h[0][i]++;
  }
  for (int i = 1; i <= l2; i++) {
    c[1][i] = c[1][i - 1];
    h[1][i] = h[1][i - 1];
    if (b[i] != 'A')
      c[1][i]++, h[1][i] = 0;
    else
      h[1][i]++;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int L1, R1, L2, R2, sum1, sum2;
    scanf("%d%d", &L1, &R1);
    sum1 = c[0][R1] - c[0][L1 - 1];
    scanf("%d%d", &L2, &R2);
    sum2 = c[1][R2] - c[1][L2 - 1];
    int d = min(h[0][R1], R1 - L1 + 1) - min(h[1][R2], R2 - L2 + 1);
    int d1 = sum2 - sum1;
    if (d1 < 0 || d1 % 2 || (d == 0 && sum1 == 0 && sum2) || d < 0)
      printf("0");
    else if (d == 0 || d1 > 0 || d % 3 == 0)
      printf("1");
    else
      printf("0");
  }
}
