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[100010], t[100010];
int s1[100010], vt[100010], s2[100010], at[100010], q;
int main() {
  scanf("%s%s", s + 1, t + 1);
  for (int i = 1; s[i] != '\0'; ++i) {
    s1[i] = s1[i - 1] + (s[i] == 'B' || s[i] == 'C');
    s2[i] = s[i] == 'A' ? s2[i - 1] + 1 : 0;
  }
  for (int i = 1; t[i] != '\0'; ++i) {
    vt[i] = vt[i - 1] + (t[i] == 'B' || t[i] == 'C');
    at[i] = t[i] == 'A' ? at[i - 1] + 1 : 0;
  }
  scanf("%d", &q);
  while (q--) {
    int a, b, c, d, s, t;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    s = s1[b] - s1[a - 1];
    t = vt[d] - vt[c - 1];
    if ((t > s) && (((t - s) & 1) == 0)) {
      s = min(s2[b], b - a);
      t = min(at[d], d - c + 1);
      if (s >= t)
        printf("1");
      else
        printf("0");
    } else if (s == t) {
      s = min(s2[b], b - a + 1);
      t = min(at[d], d - c + 1);
      if ((s >= t) && (((s - t) % 3) == 0))
        printf("1");
      else
        printf("0");
    } else
      printf("0");
  }
  puts("");
  return 0;
}
