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 N = 100005;
char s[N], t[N];
int n, m;
int ps[N], pt[N];
int ls[N], lt[N];
int main() {
  scanf("%s %s", s + 1, t + 1);
  n = (int)strlen(s + 1), m = (int)strlen(t + 1);
  for (int i = 1, sz = 0; i <= n; i++) {
    ps[i] = ps[i - 1] + int(s[i] != 'A');
    if (s[i] == 'A')
      ls[i] = ++sz;
    else
      ls[i] = sz = 0;
  }
  for (int i = 1, sz = 0; i <= m; i++) {
    pt[i] = pt[i - 1] + int(t[i] != 'A');
    if (t[i] == 'A')
      lt[i] = ++sz;
    else
      lt[i] = sz = 0;
  }
  auto checkA = [](int source, int target) {
    if (source < 0 or target < 0) return false;
    return target <= source and source % 3 == target % 3;
  };
  int q;
  scanf("%d", &q);
  while (q--) {
    int l1, r1, l2, r2;
    scanf("%d %d %d %d", &l1, &r1, &l2, &r2);
    int cnt1 = ps[r1] - ps[l1 - 1];
    int cnt2 = pt[r2] - pt[l2 - 1];
    if (!cnt1 and !cnt2) {
      printf("%d", checkA(r1 - l1 + 1, r2 - l2 + 1));
      continue;
    }
    if (!cnt2) {
      printf("0");
      continue;
    }
    if (!cnt1) {
      printf("%d", cnt2 % 2 == 0 and r1 - l1 >= lt[r2]);
      continue;
    }
    if (cnt2 >= cnt1 and cnt2 % 2 == cnt1 % 2) {
      if (checkA(ls[r1], lt[r2])) {
        printf("1");
      } else if (cnt2 >= cnt1 + 2 and ls[r1] - 1 >= lt[r2]) {
        printf("1");
      } else if (cnt2 >= cnt1 + 4 and checkA(ls[r1] - 2, lt[r2])) {
        printf("1");
      } else {
        printf("0");
      }
    } else {
      printf("0");
    }
  }
  printf("\n");
}
