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;
void calc(int *kol, int *as, string &s) {
  for (int i = 1; i <= s.size(); ++i) {
    kol[i] = kol[i - 1] + (s[i - 1] != 'A');
    if (s[i - 1] == 'A') as[i] = as[i - 1] + 1;
  }
}
int get_kol(int *kol, int a, int b) { return kol[b] - kol[a - 1]; }
const int N = 1e+5 + 20;
int ks[N], as[N];
int kt[N], at[N];
int main() {
  ios_base::sync_with_stdio(0);
  string s, t;
  cin >> s >> t;
  calc(ks, as, s);
  calc(kt, at, t);
  int q;
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int tk = get_kol(kt, c, d);
    int sk = get_kol(ks, a, b);
    int ta = min(at[d], d - c + 1);
    int sa = min(as[b], b - a + 1);
    if (tk < sk || (tk - sk) % 2 == 1 || sa < ta) {
      cout << 0;
    } else if (sk == tk) {
      cout << ((sa - ta) % 3 == 0);
    } else {
      if (sk == 0 && tk != 0) {
        sa--;
      }
      if (sa < ta) {
        cout << 0;
      } else {
        cout << 1;
      }
    }
  }
  return 0;
}
