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;
void f(string &s, vector<int> &last, vector<int> &sums, int len) {
  sums[0] = 0;
  last[0] = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] == 'A') {
      sums[i + 1] = sums[i];
      last[i + 1] = last[i] + 1;
    } else {
      sums[i + 1] = sums[i] + 1;
      last[i + 1] = 0;
    }
  }
}
bool check(vector<int> &sums1, vector<int> &last1, vector<int> &sums2,
           vector<int> &last2, int l1, int r1, int l2, int r2) {
  int a1, b1, a2, b2;
  a1 = min(last1[r1], r1 - l1);
  a2 = min(last2[r2], r2 - l2);
  b1 = sums1[r1] - sums1[l1];
  b2 = sums2[r2] - sums2[l2];
  if (a2 > a1) {
    return false;
  }
  if ((a1 - a2) % 3 != 0) {
    b1 += 2;
  }
  if (b1 == 0 && b2 > 0 && a1 == a2) {
    return false;
  }
  if (b1 > b2 || (b2 - b1) % 2 == 1) {
    return false;
  }
  return true;
}
int main() {
  int n, len1, len2, l1, r1, l2, r2;
  string s, t;
  cin >> s >> t >> n;
  len1 = s.size();
  len2 = t.size();
  vector<int> last1(len1 + 1);
  vector<int> sums1(len1 + 1);
  f(s, last1, sums1, len1);
  vector<int> last2(len2 + 1);
  vector<int> sums2(len2 + 1);
  f(t, last2, sums2, len2);
  vector<bool> res(n);
  for (int i = 0; i < n; i++) {
    cin >> l1 >> r1 >> l2 >> r2;
    l1--;
    l2--;
    res[i] = check(sums1, last1, sums2, last2, l1, r1, l2, r2);
  }
  for (int i = 0; i < n; i++) {
    cout << res[i];
  }
  cout << '\n';
  return 0;
}
