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;
const int M = 1e5 + 5, INF = 0x3f3f3f3f;
int sum[M], sum1[M];
int Sa[M], Ta[M];
char A[M], B[M];
int main() {
  scanf("%s%s", A + 1, B + 1);
  int n = strlen(A + 1), m = strlen(B + 1);
  for (int j = 1; j <= n; j++) {
    if (A[j] == 'A') {
      Sa[j] = Sa[j - 1] + 1;
      sum[j] = sum[j - 1];
    } else {
      Sa[j] = 0;
      sum[j] = sum[j - 1] + 1;
    }
  }
  for (int j = 1; j <= m; j++) {
    if (B[j] == 'A') {
      Ta[j] = Ta[j - 1] + 1;
      sum1[j] = sum1[j - 1];
    } else {
      Ta[j] = 0;
      sum1[j] = sum1[j - 1] + 1;
    }
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int x = (sum[b] - sum[a - 1]) & 1, y = (sum1[d] - sum1[c - 1]) & 1;
    int Sb = (sum[b] - sum[a - 1]), Tb = (sum1[d] - sum1[c - 1]);
    int SA = min(Sa[b], b - a + 1), TA = min(Ta[d], d - c + 1);
    if ((SA - TA) % 3) Sb += 2;
    if (Sb > Tb || SA < TA)
      printf("0");
    else if (SA == TA && Sb == 0 && Tb)
      printf("0");
    else if (x == y)
      printf("1");
    else
      printf("0");
  }
  printf("\n");
  return 0;
}
