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;
int main() {
  char S[100009], T[100009];
  int q, a, b, c, d, i, aa, bb, sa, sb, ta, tb, SA[100009], SB[100009],
      TA[100009], TB[100009];
  for (i = 1, SA[0] = SB[0] = TA[0] = TB[0] = 0,
      scanf("%s%s%d", S + 1, T + 1, &q);
       S[i]; ++i)
    SA[i] = S[i] == 'A' ? SA[i - 1] + 1 : 0, SB[i] = SB[i - 1] + (S[i] != 'A');
  for (i = 1; T[i]; ++i)
    TA[i] = T[i] == 'A' ? TA[i - 1] + 1 : 0, TB[i] = TB[i - 1] + (T[i] != 'A');
  while (q--) {
    scanf("%d%d%d%d", &a, &b, &c, &d),
        sa = min(SA[b], b - a + 1), ta = min(TA[d], d - c + 1), aa = ta - sa,
        sb = SB[b] - SB[a - 1], tb = TB[d] - TB[c - 1], bb = tb - sb;
    if (1 & bb || bb < 0 || aa > 0 || tb && !sb && !aa)
      printf("0");
    else if (bb)
      printf("1");
    else if (aa % 3 == 0)
      printf("1");
    else
      printf("0");
  }
  exit(0);
}
