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() {
  int a1[100005], a2[100005], b1[100005], b2[100005], q, l1, l2, r1, r2, as, at,
      bs, bt;
  string s, t;
  cin >> s >> t;
  s = "X" + s;
  t = "X" + t;
  for (int i = 1; i < s.length(); i++)
    if (s[i] == 'C') s[i] = 'B';
  for (int i = 1; i < t.length(); i++)
    if (t[i] == 'C') t[i] = 'B';
  a1[0] = b1[0] = 0;
  for (int i = 1; i < s.length(); i++) {
    if (s[i] == 'B') {
      b1[i] = b1[i - 1] + 1;
      a1[i] = i;
    } else {
      b1[i] = b1[i - 1];
      a1[i] = a1[i - 1];
    }
  }
  a2[0] = b2[0] = 0;
  for (int i = 1; i < t.length(); i++) {
    if (t[i] == 'B') {
      b2[i] = b2[i - 1] + 1;
      a2[i] = i;
    } else {
      b2[i] = b2[i - 1];
      a2[i] = a2[i - 1];
    }
  }
  cin >> q;
  while (q--) {
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    bs = b1[r1] - b1[l1 - 1];
    bt = b2[r2] - b2[l2 - 1];
    as = r1 - (a1[r1] < l1 ? l1 - 1 : a1[r1]);
    at = r2 - (a2[r2] < l2 ? l2 - 1 : a2[r2]);
    if (bs == bt) {
      if (at <= as && ((as - at) % 3 == 0))
        printf("1");
      else
        printf("0");
    } else if (bs < bt) {
      if ((bt - bs) % 2 == 0) {
        if (bs) {
          if (at <= as)
            printf("1");
          else
            printf("0");
        } else {
          if (at <= (as - 1))
            printf("1");
          else
            printf("0");
        }
      } else
        printf("0");
    } else
      printf("0");
  }
  return 0;
}
