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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  out << p.first << " " << p.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& arr) {
  for (auto& i : arr) {
    in >> i;
  }
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> arr) {
  for (int i = 0; i < (int)arr.size() - 1; ++i) out << arr[i] << " ";
  if (arr.size()) out << arr.back();
  out << '\n';
  return out;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  vector<bool> a(n);
  vector<bool> b(m);
  for (int i = 0; i < s.size(); ++i) {
    a[i] = (s[i] == 'A');
  }
  for (int i = 0; i < t.size(); ++i) {
    b[i] = (t[i] == 'A');
  }
  vector<int> prefa(n + 1), prefb(m + 1);
  for (int i = 0; i < n; ++i) {
    prefa[i + 1] = prefa[i] + !a[i];
  }
  for (int i = 0; i < m; ++i) {
    prefb[i + 1] = prefb[i] + !b[i];
  }
  vector<int> lena(n + 1), lenb(m + 1);
  for (int i = 0; i < n; ++i) {
    if (a[i]) {
      lena[i + 1] = lena[i] + 1;
    }
  }
  for (int i = 0; i < m; ++i) {
    if (b[i]) {
      lenb[i + 1] = lenb[i] + 1;
    }
  }
  int q;
  cin >> q;
  vector<int> ans;
  for (int it = 0; it < q; ++it) {
    int x, y, c, d;
    cin >> x >> y >> c >> d;
    int cnta = prefa[y] - prefa[x - 1];
    int cntb = prefb[d] - prefb[c - 1];
    int lna = min(lena[y], y - x + 1);
    int lnb = min(lenb[d], d - c + 1);
    if (lna < lnb) {
      ans.push_back(0);
      continue;
    }
    if (cnta == cntb && (lna - lnb) % 3 == 0) {
      ans.push_back(1);
      continue;
    }
    if (lna != lnb) {
      cnta += 2;
    }
    if (cnta > cntb) {
      ans.push_back(0);
      continue;
    }
    if (cnta == cntb) {
      ans.push_back(1);
      continue;
    }
    if (cnta == 0) {
      ans.push_back(0);
      continue;
    }
    if ((cntb - cnta) % 2 == 0) {
      ans.push_back(1);
    } else {
      ans.push_back(0);
    }
  }
  for (int i = 0; i < q; ++i) cout << ans[i];
  cout << '\n';
  return 0;
}
