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 int maxn = 1e5 + 10;
int n, m, q;
string s, t;
int sum_s[maxn], sum_t[maxn], a_s[maxn], a_t[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> s >> t;
  n = s.size();
  m = t.size();
  for (int i = 0; i < n; i++) {
    sum_s[i + 1] = sum_s[i] + (s[i] != 'A');
    if (s[i] == 'A') {
      a_s[i + 1] = a_s[i] + 1;
    } else {
      a_s[i + 1] = 0;
    }
  }
  for (int i = 0; i < m; i++) {
    sum_t[i + 1] = sum_t[i] + (t[i] != 'A');
    if (t[i] == 'A') {
      a_t[i + 1] = a_t[i] + 1;
    } else {
      a_t[i + 1] = 0;
    }
  }
  cin >> q;
  for (int i = 0, a, b, c, d, ss, st, ls, lt, sfs, sft; i < q; i++) {
    cin >> a >> b >> c >> d;
    ss = (sum_s[b] - sum_s[a - 1]);
    st = (sum_t[d] - sum_t[c - 1]);
    ls = b - a + 1;
    lt = d - c + 1;
    sfs = min(a_s[b], ls);
    sft = min(a_t[d], lt);
    if (ss % 2 != st % 2) {
      cout << "0";
    } else {
      if (ss > st) {
        cout << "0";
      } else if (ss == st) {
        if (sfs >= sft && sfs % 3 == sft % 3) {
          cout << "1";
        } else {
          cout << "0";
        }
      } else {
        if (ss == 0 && st > 0 && sfs == sft) {
          cout << "0";
        } else if (sfs >= sft) {
          cout << "1";
        } else {
          cout << "0";
        }
      }
    }
  }
  cout << "\n";
  return 0;
}
