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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s;
  string t;
  cin >> s >> t;
  for (long long i = 0; i < s.size(); i++) {
    if (s[i] == 'C') s[i] = 'B';
  }
  for (long long i = 0; i < t.size(); i++) {
    if (t[i] == 'C') t[i] = 'B';
  }
  long long n = s.size(), m = t.size();
  long long prefs[n + 1], preft[m + 1];
  prefs[0] = 0, preft[0] = 0;
  for (long long i = 0; i < n; i++) {
    prefs[i + 1] = prefs[i] + (s[i] == 'B');
  }
  for (long long i = 0; i < m; i++) {
    preft[i + 1] = preft[i] + (t[i] == 'B');
  }
  long long ls[n + 1], lt[m + 1];
  ls[0] = 0, lt[0] = 0;
  for (long long i = 0; i < n; i++) {
    if (s[i] == 'B')
      ls[i + 1] = 0;
    else
      ls[i + 1] = ls[i] + 1;
  }
  for (long long i = 0; i < m; i++) {
    if (t[i] == 'B')
      lt[i + 1] = 0;
    else
      lt[i + 1] = lt[i] + 1;
  }
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    a--, b--, c--, d--;
    long long F = prefs[b + 1] - prefs[a];
    long long S = preft[d + 1] - preft[c];
    if (F % 2 != S % 2) {
      cout << 0;
      continue;
    }
    if (F > S) {
      cout << 0;
      continue;
    }
    long long A = ls[b + 1], B = lt[d + 1];
    A = min(A, b - a + 1);
    B = min(B, d - c + 1);
    if (A < B) {
      cout << 0;
      continue;
    }
    if (F == 0) {
      if (S != 0) {
        if (A == B)
          cout << 0;
        else
          cout << 1;
      } else {
        if (A % 3 == B % 3)
          cout << 1;
        else
          cout << 0;
      }
      continue;
    } else if (A % 3 == B % 3)
      cout << 1;
    else {
      F += 2;
      if (F > S)
        cout << 0;
      else
        cout << 1;
    }
  }
  return 0;
}
