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>
const int N = 100005;
int l, r, suf1[N], suf2[N], pre1[N], pre2[N], n, m, Q;
char s[N], t[N];
int solve() {
  scanf("%d%d", &l, &r);
  int cnt1 = std::min(r - l + 1, suf1[r]), s1 = pre1[r] - pre1[l - 1];
  scanf("%d%d", &l, &r);
  int cnt2 = std::min(r - l + 1, suf2[r]), s2 = pre2[r] - pre2[l - 1];
  if (s1 > s2) return 0;
  if ((s2 - s1) & 1) return 0;
  if (cnt1 < cnt2) return 0;
  if (s1 == 0 && s2 && cnt1 == cnt2) return 0;
  if (s1 == s2 && (cnt2 - cnt1) % 3) return 0;
  return 1;
}
int main() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 1; i <= n; i++)
    if (s[i] == 'A') suf1[i] = suf1[i - 1] + 1;
  for (int i = 1; i <= m; i++)
    if (t[i] == 'A') suf2[i] = suf2[i - 1] + 1;
  for (int i = 1; i <= n; i++) pre1[i] = pre1[i - 1] + (s[i] != 'A');
  for (int i = 1; i <= m; i++) pre2[i] = pre2[i - 1] + (t[i] != 'A');
  scanf("%d", &Q);
  while (Q--) printf("%d", solve());
}
