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;
const long long inf = 1000000007;
int ta[(1 << 20)], tb[(1 << 20)], za[(1 << 20)], zb[(1 << 20)];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string A, B;
  cin >> A >> B;
  int n = A.size(), m = B.size();
  A = " " + A;
  B = " " + B;
  int c = 0;
  for (int i = 1; i <= n; ++i) {
    if (A[i] != 'A')
      ++ta[i], c = 0;
    else
      ++c;
    ta[i] += ta[i - 1];
    za[i] = c;
  }
  c = 0;
  for (int i = 1; i <= m; ++i) {
    if (B[i] != 'A')
      ++tb[i], c = 0;
    else
      ++c;
    zb[i] = c;
    tb[i] += tb[i - 1];
  }
  int Q;
  cin >> Q;
  while (Q-- > 0) {
    int a1, a2, b1, b2;
    cin >> a1 >> b1 >> a2 >> b2;
    int c1 = ta[b1] - ta[a1 - 1], c2 = tb[b2] - tb[a2 - 1];
    int aa = min(za[b1], b1 - a1 + 1);
    int ba = min(zb[b2], b2 - a2 + 1);
    if (c1 > c2)
      cout << 0;
    else if (c1 == c2) {
      if (aa >= ba && (aa - ba) % 3 == 0)
        cout << 1;
      else
        cout << 0;
    } else {
      if ((c2 - c1) % 2 == 1)
        cout << 0;
      else if (ba > aa)
        cout << 0;
      else if (ba == aa && aa == b1 - a1 + 1)
        cout << 0;
      else
        cout << 1;
    }
  }
}
