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;
using lint = long long;
template <typename T>
using vc = std::vector<T>;
void solve(istream& cin, ostream& cout) {
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  vector<int> ss(n), st(m);
  vector<int> ls(n), lt(m);
  for (int i = (0); i < int(n); ++i) {
    if (s[i] != 'A') {
      ss[i] = 1;
    } else {
      ls[i] = 1;
      if (i) {
        ls[i] += ls[i - 1];
      }
    }
    if (i) {
      ss[i] += ss[i - 1];
    }
  }
  for (int i = (0); i < int(m); ++i) {
    if (t[i] != 'A') {
      st[i] = 1;
    } else {
      lt[i] = 1;
      if (i) {
        lt[i] += lt[i - 1];
      }
    }
    if (i) {
      st[i] += st[i - 1];
    }
  }
  int q;
  cin >> q;
  for (int iq = (0); iq < int(q); ++iq) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    --a;
    --b;
    --c;
    --d;
    int x1 = ss[b];
    if (a) {
      x1 -= ss[a - 1];
    }
    int x2 = st[d];
    if (c) {
      x2 -= st[c - 1];
    }
    int l1 = min(ls[b], b - a + 1);
    int l2 = min(lt[d], d - c + 1);
    if (x1 > x2) {
      cout << 0;
      continue;
    }
    if (l1 < l2) {
      cout << 0;
      continue;
    }
    if (x1 == x2) {
      if (l1 % 3 != l2 % 3) {
        cout << 0;
        continue;
      }
      cout << 1;
      continue;
    }
    if (x1 == 0 && l1 == l2 && l1 != 0) {
      cout << 0;
      continue;
    }
    if (x2 % 2 != x1 % 2) {
      cout << 0;
      continue;
    }
    cout << 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve(cin, cout);
  return 0;
}
