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;
bool f(array<int, 2> p1, array<int, 2> p2) {
  if (p1[0] > p2[0] || ((p1[0] ^ p2[0]) & 1) || p1[1] < p2[1]) return false;
  if (p1[0] == p2[0]) return (p1[1] - p2[1]) % 3 == 0;
  if (!p1[0]) {
    return f({2, p1[1] - 1}, p2) || (p1[1] >= 2 && f({2, p1[1] - 2}, p2)) ||
           (p1[1] >= 3 && f({2, p1[1] - 3}, p2));
  }
  return true;
}
struct DS {
  string s;
  vector<int> cs, last;
  DS() {
    cin >> s;
    cs.resize(s.size());
    last.resize(s.size());
    for (int i = 0; i < s.size(); ++i) {
      cs[i] = (i ? cs[i - 1] : 0) + (s[i] != 'A');
      last[i] = s[i] != 'A' ? i : i ? last[i - 1] : -1;
    }
  }
  array<int, 2> query(int i, int j) {
    if (last[j] < i) return {0, j - i + 1};
    return {cs[last[j]] - (i ? cs[i - 1] : 0), j - last[j]};
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  DS ds1;
  DS ds2;
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    --a;
    --b;
    --c;
    --d;
    cout << f(ds1.query(a, b), ds2.query(c, d));
  }
  return 0;
}
