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 cbs[N], cbt[N];
int pbs[N], pbt[N];
void go(char *s, int *cntB, int *prvB) {
  prvB[0] = 0;
  for (int i = 1; s[i]; i++) {
    cntB[i] = cntB[i - 1] + (s[i] != 'A');
    prvB[i] = (s[i] != 'A' ? i : prvB[i - 1]);
  }
}
int main() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  go(s, cbs, pbs);
  go(t, cbt, pbt);
  int q, a, b, c, d;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int counts = cbs[b] - cbs[a - 1];
    int countt = cbt[d] - cbt[c - 1];
    int sufs = b - max(pbs[b], a - 1);
    int suft = d - max(pbt[d], c - 1);
    if (counts > countt || (countt - counts) % 2)
      putc('0', stdout);
    else if (counts == 0 && countt != 0 && (sufs <= suft))
      putc('0', stdout);
    else if (counts == countt && ((sufs < suft) || (sufs - suft) % 3))
      putc('0', stdout);
    else if (counts < countt && (sufs < suft))
      putc('0', stdout);
    else
      putc('1', stdout);
  }
  puts("");
  return 0;
}
