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;
const int mxN = 1e5;
int n, m, q, p1[mxN + 1], p2[mxN + 1], prv1[mxN + 1], prv2[mxN + 1];
string s, t;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> t >> q;
  int n = s.size(), m = t.size();
  for (int i = 0; i < n; ++i) {
    p1[i + 1] = p1[i] + (s[i] != 'A');
    prv1[i + 1] = s[i] == 'A' ? 1 + prv1[i] : 0;
  }
  for (int i = 0; i < m; ++i) {
    p2[i + 1] = p2[i] + (t[i] != 'A');
    prv2[i + 1] = t[i] == 'A' ? 1 + prv2[i] : 0;
  }
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d, --a, --b, --c, --d;
    int cnt1 = p1[b + 1] - p1[a];
    int cnt2 = p2[d + 1] - p2[c];
    int a1 = min(b - a + 1, prv1[b + 1]);
    int a2 = min(d - c + 1, prv2[d + 1]);
    if (cnt1 > cnt2 || (cnt2 - cnt1) % 2 || a1 < a2) {
      cout << 0;
      continue;
    }
    int need = (cnt2 - cnt1) / 2;
    if (a1 == a2 && !cnt1 && need || !need && (a1 - a2) % 3) {
      cout << 0;
      continue;
    }
    cout << 1;
  }
  return 0;
}
