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;
struct psum {
  int n;
  vector<int> b;
  vector<int> a;
  psum(string s) : n(s.size()) {
    b.assign(n, 0);
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      cnt += s[i] == 'B';
      b[i] = cnt;
    }
    a.assign(n, 0);
    cnt = 0;
    for (int i = 0; i < n; i++) {
      if (s[i] == 'A') {
        cnt++;
      } else {
        cnt = 0;
      }
      a[i] = cnt;
    }
  }
  int get(int l, int r) { return b[r] - (l == 0 ? 0 : b[l - 1]); }
};
int main() {
  ios_base::sync_with_stdio(0);
  string s;
  cin >> s;
  replace(s.begin(), s.end(), 'C', 'B');
  string t;
  cin >> t;
  replace(t.begin(), t.end(), 'C', 'B');
  psum cs(s);
  psum ct(t);
  int z;
  cin >> z;
  while (z--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    --a;
    --b;
    --c;
    --d;
    int ws = cs.get(a, b);
    int wt = ct.get(c, d);
    int as = min(b - a + 1, cs.a[b]);
    int at = min(d - c + 1, ct.a[d]);
    if (ws > wt) {
      cout << "0";
      continue;
    }
    if ((wt - ws) % 2 == 1) {
      cout << "0";
      continue;
    }
    if (at > as) {
      cout << "0";
      continue;
    }
    if (ws + 2 <= wt && at < as) {
      cout << "1";
      continue;
    }
    if (wt != 0 && ws == 0) {
      ws += 2;
      --as;
    }
    if ((as - at) % 3 == 0) {
      cout << "1";
      continue;
    }
    cout << "0";
    continue;
  }
}
