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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string s, t;
  cin >> s >> t;
  int n = s.size(), m = t.size();
  vector<int> bs(n), bt(m);
  vector<int> as(n), at(m);
  int sum = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] != 'A') sum++;
    bs[i] = sum;
    as[i] = (i == 0) ? 0 : as[i - 1];
    if (s[i] == 'A')
      as[i]++;
    else
      as[i] = 0;
  }
  sum = 0;
  for (int i = 0; i < m; i++) {
    if (t[i] != 'A') sum++;
    bt[i] = sum;
    at[i] = (i == 0) ? 0 : at[i - 1];
    if (t[i] == 'A')
      at[i]++;
    else
      at[i] = 0;
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l0, r0, l1, r1;
    cin >> l0 >> r0 >> l1 >> r1;
    l0--, r0--, l1--, r1--;
    int s0 = bs[r0] - ((l0 == 0) ? 0 : bs[l0 - 1]),
        t0 = bt[r1] - ((l1 == 0) ? 0 : bt[l1 - 1]);
    int s1 = as[r0], t1 = at[r1];
    if (s1 > (r0 - l0 + 1)) s1 = r0 - l0 + 1;
    if (t1 > (r1 - l1 + 1)) t1 = r1 - l1 + 1;
    if (s1 == t1 && s0 == 0 && t0 > 0) {
      cout << 0;
    } else if ((s0 <= t0) && (s1 >= t1)) {
      if ((s1 - t1) % 3 > 0) s0 += 2;
      if ((s0 <= t0) && (t0 - s0) % 2 == 0)
        cout << 1;
      else
        cout << 0;
    } else
      cout << 0;
  }
  cout << "\n";
  return 0;
}
