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(0);
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  for (int i = 0; i < n; i++) {
    if (s[i] == 'C') s[i] = 'B';
  }
  for (int i = 0; i < m; i++) {
    if (t[i] == 'C') t[i] = 'B';
  }
  vector<int> as(n + 1), bs(n + 1), at(m + 1), bt(m + 1);
  for (int i = 0; i < n; i++) {
    if (s[i] == 'A')
      as[i + 1] = as[i] + 1;
    else
      bs[i + 1]++;
  }
  for (int i = 0; i < m; i++) {
    if (t[i] == 'A')
      at[i + 1] = at[i] + 1;
    else
      bt[i + 1]++;
  }
  for (int i = 1; i <= n; i++) {
    bs[i] += bs[i - 1];
  }
  for (int i = 1; i <= m; i++) {
    bt[i] += bt[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a--;
    b--;
    c--;
    d--;
    int bsk = bs[b + 1] - bs[a];
    int btk = bt[d + 1] - bt[c];
    int ask = min(as[b + 1], b - a + 1);
    int atk = min(at[d + 1], d - c + 1);
    if (ask > atk && ask % 3 != atk % 3) bsk += 2;
    if (bsk == 0 && btk > 0 && ask == atk)
      cout << 0;
    else if (ask >= atk && bsk % 2 == btk % 2 && bsk <= btk)
      cout << 1;
    else
      cout << 0;
  }
}
