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 pre[2][100005];
int bs(int k, int n) {
  int l = 1, r = 100000, ret = -1;
  while (l <= r) {
    int mid = l + r >> 1;
    if (pre[k][mid] >= n) {
      r = mid - 1;
    } else {
      l = mid + 1;
      ret = mid;
    }
  }
  ret++;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int k = 0; k < 2; k++) {
    string s;
    cin >> s;
    for (int n = 1; n <= s.size(); n++) {
      if (s[n - 1] != 'A') {
        pre[k][n]++;
      }
    }
    partial_sum(pre[k], pre[k] + 100001, pre[k]);
  }
  int Q;
  cin >> Q;
  while (Q--) {
    int l1, r1, l0, r0;
    cin >> l0 >> r0 >> l1 >> r1;
    int f0 = pre[0][r0] - pre[0][l0 - 1];
    int f1 = pre[1][r1] - pre[1][l1 - 1];
    bool cond1 = f1 >= f0 && (f1 % 2 == f0 % 2);
    int z0 =
        max(l0 - 1,
            (int)(lower_bound(pre[0], pre[0] + 100001, pre[0][r0]) - pre[0]));
    int z1 =
        max(l1 - 1,
            (int)(lower_bound(pre[1], pre[1] + 100001, pre[1][r1]) - pre[1]));
    z0 = r0 - z0;
    z1 = r1 - z1;
    bool cond2 = (z0 >= z1 && ((z0 - z1) % 3 != 0) * 2 + f0 <= f1) &&
                 (z0 != z1 || !!f1 == !!f0);
    if (cond1 && cond2) {
      cout << 1;
    } else {
      cout << 0;
    }
  }
}
