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 pii = std::pair<int, int>;
using pll = std::pair<long long, long long>;
const int maxn = 1e5 + 100;
struct string {
  std::string data;
  int prefix[maxn];
  int a_before[maxn];
  inline int query_range(int l, int r) { return prefix[r] - prefix[l - 1]; }
};
std::istream& operator>>(std::istream& in, string& s) {
  in >> s.data;
  for (int i = 1; i <= (int)s.data.length(); i++) {
    s.prefix[i] = s.prefix[i - 1] + (s.data[i - 1] != 'A');
  }
  for (int i = 1; i <= (int)s.data.length(); i++) {
    if (s.data[i - 1] == 'A')
      s.a_before[i] = s.a_before[i - 1] + 1;
    else
      s.a_before[i] = 0;
  }
  return in;
}
string s, t;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin >> s >> t;
  int q;
  std::cin >> q;
  while (q--) {
    int a, b, c, d;
    std::cin >> a >> b >> c >> d;
    int c1 = s.query_range(a, b);
    int c2 = t.query_range(c, d);
    int ca1 = s.a_before[b];
    int ca2 = t.a_before[d];
    ca1 = std::min(ca1, b - a + 1);
    ca2 = std::min(ca2, d - c + 1);
    if (c1 == c2) {
      if (ca2 <= ca1 && (ca1 - ca2) % 3 == 0)
        std::cout << 1;
      else
        std::cout << 0;
    } else if (c1 < c2 && (c2 - c1) % 2 == 0 && ca1 >= ca2) {
      if (ca1 == b - a + 1 && ca1 == ca2) {
        std::cout << 0;
      } else {
        std::cout << 1;
      }
    } else {
      std::cout << 0;
    }
  }
  return 0;
}
