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 MX = 1e5 + 5;
char s[MX], t[MX];
int sum1[MX], sum2[MX];
int pre1[MX], pre2[MX];
int main() {
  scanf("%s%s", s + 1, t + 1);
  int n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 1; i <= n; i++) {
    sum1[i] = sum1[i - 1] + (s[i] != 'A');
    pre1[i] = s[i] == 'A' ? pre1[i - 1] + 1 : 0;
  }
  for (int i = 1; i <= m; i++) {
    sum2[i] = sum2[i - 1] + (t[i] != 'A');
    pre2[i] = t[i] == 'A' ? pre2[i - 1] + 1 : 0;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int x = sum2[d] - sum2[c - 1] - sum1[b] + sum1[a - 1];
    int y = min(b - a + 1, pre1[b]) - min(d - c + 1, pre2[d]);
    if (x < 0 || x % 2 == 1 || y < 0 ||
        y == 0 && x > 0 && sum1[b] - sum1[a - 1] == 0) {
      printf("0");
      continue;
    }
    if (y == 0 || x > 0 || y % 3 == 0)
      printf("1");
    else
      printf("0");
  }
  return 0;
}
