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;
class StringInfGetter {
 public:
  string &s;
  vector<int> a, b;
  int n;
  StringInfGetter(string &s1) : s(s1) {
    n = s.size();
    a.assign(n + 1, 0);
    b.assign(n + 1, 0);
    int anum = 0;
    for (int i = 0; i < int(n); i++) {
      anum = (s[i] == 'A' ? anum + 1 : 0);
      b[i + 1] = b[i] + (s[i] == 'B' || s[i] == 'C');
      a[i + 1] = anum;
    }
  }
  pair<int, int> getInf(int l, int r) {
    return make_pair((min(r - l + 1, a[r])), (b[r] - b[l - 1]));
  }
};
bool canConst(pair<int, int> a, pair<int, int> b) {
  if (a == b) return 1;
  if (b.second < a.second || ((b.second - a.second) % 2)) return 0;
  if (b.second > a.second)
    return (a.second ? a.first >= b.first : a.first > b.first);
  if (b.second == a.second)
    return a.first >= b.first && ((a.first - b.first) % 3) == 0;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  string s, t;
  cin >> s >> t;
  StringInfGetter a(s), b(t);
  int n;
  cin >> n;
  for (int i = 0; i < int(n); i++) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    pair<int, int> p = a.getInf(l1, r1);
    pair<int, int> q = b.getInf(l2, r2);
    cout << canConst(p, q);
  }
}
