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;
int main() {
  string s1, s2;
  cin >> s1 >> s2;
  for (int i = 0; i < s1.length(); ++i) {
    if (s1[i] == 'C') {
      s1[i] = 'B';
    }
  }
  for (int i = 0; i < s2.length(); ++i) {
    if (s2[i] == 'C') {
      s2[i] = 'B';
    }
  }
  vector<int> b1(s1.length() + 1), b2(s2.length() + 1);
  vector<int> a1(s1.length() + 1), a2(s2.length() + 1);
  for (int i = 1; i < s1.length() + 1; ++i) {
    b1[i] = b1[i - 1] + (s1[i - 1] == 'B');
    a1[i] = (a1[i - 1] + 1) * (s1[i - 1] == 'A');
  }
  for (int i = 1; i < s2.length() + 1; ++i) {
    b2[i] = b2[i - 1] + (s2[i - 1] == 'B');
    a2[i] = (a2[i - 1] + 1) * (s2[i - 1] == 'A');
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int l1, l2, r1, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    int cnt1 = b1[r1] - b1[l1 - 1], cnt2 = b2[r2] - b2[l2 - 1];
    int cnta1 = min(r1 - l1 + 1, a1[r1]), cnta2 = min(r2 - l2 + 1, a2[r2]);
    if (cnta1 < cnta2) {
      cout << 0;
      continue;
    }
    if (cnta1 == cnta2 && (cnt1 == 0 && cnt2 > 0)) {
      cout << 0;
      continue;
    }
    if (cnta1 % 3 != cnta2 % 3 || (cnt1 == 0 && cnt2 > 0)) {
      cnt1 += 2;
    }
    if (cnt1 > cnt2) {
      cout << 0;
      continue;
    }
    if ((cnt2 - cnt1) % 2 != 0) {
      cout << 0;
      continue;
    }
    cout << 1;
  }
  return 0;
}
