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 maxn = 1e5;
int n, m;
char s[maxn + 3], t[maxn + 3];
int cnt1[maxn + 3], cnt2[maxn + 3];
int lst1[maxn + 3], lst2[maxn + 3];
int main(void) {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; ++i) {
    cnt1[i] = cnt1[i - 1] + (s[i] != 'A');
    lst1[i] = s[i] == 'A' ? lst1[i - 1] : i;
  }
  for (int i = 1; i <= m; ++i) {
    cnt2[i] = cnt2[i - 1] + (t[i] != 'A');
    lst2[i] = t[i] == 'A' ? lst2[i - 1] : i;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int a = r1 - max(l1 - 1, lst1[r1]), b = r2 - max(l2 - 1, lst2[r2]);
    int c = cnt1[r1] - cnt1[l1 - 1], d = cnt2[r2] - cnt2[l2 - 1];
    if (a < b) {
      putchar('0');
      continue;
    }
    if ((a - b) % 3) {
      c += 2;
    }
    if (c <= d && !(d - c & 1) && !(r1 - l1 + 1 == a && d && a == b)) {
      putchar('1');
    } else {
      putchar('0');
    }
  }
  putchar('\n');
  return 0;
}
