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;
string s, t;
int sc[100005];
int bc[100005];
int ca[100005], cb[100005];
;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> t;
  int cnt = 0;
  for (int i = 0; i < s.size(); i++) {
    sc[i + 1] = sc[i];
    if (s[i] == 'A')
      cnt++;
    else
      cnt = 0, sc[i + 1]++;
    ca[i + 1] = cnt;
  }
  cnt = 0;
  for (int i = 0; i < t.size(); i++) {
    bc[i + 1] = bc[i];
    if (t[i] == 'A')
      cnt++;
    else
      cnt = 0, bc[i + 1]++;
    cb[i + 1] = cnt;
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int k = (bc[d] - bc[c - 1]) - (sc[b] - sc[a - 1]);
    int k1 = min(ca[b], b - a + 1);
    int k2 = min(cb[d], d - c + 1);
    if (k < 0)
      cout << '0';
    else {
      if (k1 < k2)
        cout << '0';
      else {
        if (k == 0)
          cout << ((k1 - k2) % 3 == 0);
        else {
          if (k % 2 == 0 && k1 - ((sc[b] - sc[a - 1]) == 0) >= k2)
            cout << '1';
          else
            cout << '0';
        }
      }
    }
  }
  cout << '\n';
  return 0;
}
