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 n, q, i, j, k, a1[100010], b1[100010], a2[100010], b2[100010], l1, l2, r1,
    r2, A1, B1, A2, B2, ans, S1, S2;
string s1, s2;
int main() {
  cin >> s1;
  cin >> s2;
  a1[0] = 0;
  b1[0] = 0;
  a2[0] = 0;
  b2[0] = 0;
  n = s1.size();
  for (i = 1; i <= n; i++) {
    a1[i] = a1[i - 1];
    b1[i] = b1[i - 1];
    if (s1[i - 1] == 'A')
      a1[i]++;
    else {
      a1[i] = 0;
      b1[i]++;
    }
  }
  n = s2.size();
  for (i = 1; i <= n; i++) {
    a2[i] = a2[i - 1];
    b2[i] = b2[i - 1];
    if (s2[i - 1] == 'A')
      a2[i]++;
    else {
      a2[i] = 0;
      b2[i]++;
    }
  }
  for (cin >> q; q > 0; q--) {
    ans = 1;
    cin >> l1 >> r1 >> l2 >> r2;
    S1 = r1 - l1 + 1;
    S2 = r2 - l2 + 1;
    A1 = min(a1[r1], S1);
    A2 = min(a2[r2], S2);
    if (S1 == A1 && S2 == A2) {
      A1 -= A2;
      if (!(A1 >= 0 && A1 % 3 == 0)) ans = 0;
      printf("%d", ans);
      continue;
    }
    if (S1 == A1) {
      B1 = 2;
      A1--;
      if (A1 >= A2)
        A1 = A2;
      else
        A1 = 0;
    } else
      B1 = b1[r1] - b1[l1 - 1];
    A1 -= A2;
    B2 = b2[r2] - b2[l2 - 1];
    if (A1 < 0) ans = 0;
    A1 = A1 % 3;
    if (A1 > 0) B1 += 2;
    if (!(B2 >= B1 && (B2 - B1) % 2 == 0)) ans = 0;
    printf("%d", ans);
  }
  printf("\n");
  return 0;
}
