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>
char a[100050], b[100050];
int sa[100050], sb[100050], ta[100050], tb[100050], Q;
int main() {
  scanf("%s%s", a + 1, b + 1);
  int la = strlen(a + 1), lb = strlen(b + 1);
  for (int i = 1; i <= la; i++) {
    if (a[i] == 'A') sa[i] = sa[i - 1] + 1;
    sb[i] = sb[i - 1] + (a[i] == 'B' || a[i] == 'C');
  }
  for (int i = 1; i <= lb; i++) {
    if (b[i] == 'A') ta[i] = ta[i - 1] + 1;
    tb[i] = tb[i - 1] + (b[i] == 'B' || b[i] == 'C');
  }
  int l, r, lsa, lta, ssb, stb;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d", &l, &r);
    lsa = std::min(r - l + 1, sa[r]), ssb = sb[r] - sb[l - 1];
    scanf("%d%d", &l, &r);
    lta = std::min(r - l + 1, ta[r]), stb = tb[r] - tb[l - 1];
    if (lsa < lta || ((ssb & 1) != (stb & 1)) || stb < ssb)
      printf("0");
    else if (lsa > lta && stb > ssb ||
             ((lsa - lta) % 3 == 0 && (ssb || (ssb == stb))))
      printf("1");
    else
      printf("0");
  }
}
