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;
vector<long long> gd(string s) {
  vector<long long> res(((long long)(s).size()) + 1, 0);
  for (long long i = 1; i < (((long long)(s).size()) + 1); ++i) {
    res[i] = res[i - 1] + (s[i - 1] == 'C' || s[i - 1] == 'B');
  }
  return res;
}
vector<long long> ga(string s) {
  vector<long long> res(((long long)(s).size()), 0);
  res[0] = s[0] == 'A';
  for (long long i = 1; i < (((long long)(s).size())); ++i) {
    if (s[i] == 'A') res[i] = res[i - 1] + 1;
  }
  return res;
}
void pv(vector<long long> s) {
  for (auto it = (s).begin(); it != (s).end(); ++it) cout << *it << " ";
  cout << endl;
}
int32_t main() {
  string s, t;
  cin >> s >> t;
  vector<long long> ds = gd(s), as = ga(s), dt = gd(t), at = ga(t);
  long long q;
  cin >> q;
  string res(q, '0');
  while (q--) {
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    a--;
    b--;
    c--;
    d--;
    long long dsc = ds[b + 1] - ds[a], dtc = dt[d + 1] - dt[c],
              atc = min(at[d], d - c + 1), asc = min(as[b], b - a + 1);
    if (atc > asc) continue;
    if ((asc - atc) % 3 != 0) dsc += 2;
    if (atc == asc && asc == b - a + 1 && atc != d - c + 1) continue;
    if (dsc > dtc || (dtc - dsc) % 2 != 0) continue;
    res[((long long)(res).size()) - 1 - q] = '1';
  }
  cout << res;
  return 0;
}
