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;
char check(int& a, int& b, int& c, int& d, vector<int>& cnt_s,
           vector<int>& cnt_t, vector<int>& con_s, vector<int>& con_t) {
  int Con_s = min(con_s[b], b - a + 1), Con_t = min(con_t[d], d - c + 1);
  int Cnt_s = cnt_s[b] - cnt_s[a - 1], Cnt_t = cnt_t[d] - cnt_t[c - 1];
  if (Con_s > Con_t) {
    if ((Con_s - Con_t) % 3) Cnt_s += 2;
  }
  if (Con_s < Con_t || Cnt_s > Cnt_t || (Cnt_t % 2 != Cnt_s % 2) ||
      (Con_s == Con_t && !Cnt_s && Cnt_t))
    return '0';
  else
    return '1';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  string s, t;
  cin >> s >> t;
  s = '*' + s;
  t = '*' + t;
  vector<int> cnt_s(s.size()), cnt_t(t.size());
  cnt_s[0] = cnt_t[0] = 0;
  for (int i = 1; i < s.size(); ++i) cnt_s[i] = cnt_s[i - 1] + (s[i] != 'A');
  for (int i = 1; i < t.size(); ++i) cnt_t[i] = cnt_t[i - 1] + (t[i] != 'A');
  vector<int> con_s(s.size()), con_t(t.size());
  con_s[1] = (s[1] == 'A');
  con_t[1] = (t[1] == 'A');
  for (int i = 2; i < s.size(); ++i)
    con_s[i] = (s[i] == 'A') ? (con_s[i - 1] + 1) : 0;
  for (int i = 2; i < t.size(); ++i)
    con_t[i] = (t[i] == 'A') ? (con_t[i - 1] + 1) : 0;
  int q;
  cin >> q;
  string ans(q, '1');
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    ans[q] = check(a, b, c, d, cnt_s, cnt_t, con_s, con_t);
  }
  reverse(ans.begin(), ans.end());
  cout << ans;
  return 0;
}
