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;
char s[100005];
int ls;
char t[100005];
int lt;
int cntas[100005];
int cntbs[100005];
int cntat[100005];
int cntbt[100005];
int q;
bool Check(int cbs, int cbt, int cas, int cat) {
  if (cbs == 0 && cbt > 0 && (cbt % 2) == 0) return cas > cat;
  if (cbs < cbt && (cbs % 2) == (cbt % 2) && cas >= cat) return true;
  if (cbs == cbt && cas >= cat && (cas % 3) == (cat % 3)) return true;
  return false;
}
int main() {
  scanf("%s%s%d", s + 1, t + 1, &q);
  ls = strlen(s + 1);
  lt = strlen(t + 1);
  for (int i = 1; i <= ls; ++i) {
    if (s[i] == 'A') cntas[i] = cntas[i - 1] + 1;
    cntbs[i] = cntbs[i - 1] + (s[i] != 'A');
  }
  for (int i = 1; i <= lt; ++i) {
    if (t[i] == 'A') cntat[i] = cntat[i - 1] + 1;
    cntbt[i] = cntbt[i - 1] + (t[i] != 'A');
  }
  while (q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    putchar('0' + Check(cntbs[b] - cntbs[a - 1], cntbt[d] - cntbt[c - 1],
                        min(b - a + 1, cntas[b]), min(d - c + 1, cntat[d])));
  }
  putchar('\n');
  return 0;
}
