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 + 14;
int n, m;
struct S {
  string s;
  int n;
  int preA[maxn], cntB[maxn];
  void build() {
    for (auto &c : s)
      if (c == 'C') c = 'B';
    n = s.size();
    for (int i = 0; i < n; i++) cntB[i + 1] = cntB[i] + (s[i] == 'B');
    for (int i = 0; i < n; i++)
      if (s[i] == 'A') preA[i + 1] = 1 + preA[i];
  }
} s, p;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> s.s >> p.s;
  s.build();
  p.build();
  int q;
  cin >> q;
  while (q--) {
    int sl, sr, pl, pr;
    cin >> sl >> sr >> pl >> pr;
    sl--, pl--;
    int sb = s.cntB[sr] - s.cntB[sl];
    if (min(sr - sl, s.preA[sr]) % 3 != min(pr - pl, p.preA[pr]) % 3) sb += 2;
    if (min(sr - sl, s.preA[sr]) < min(pr - pl, p.preA[pr]) ||
        sb > p.cntB[pr] - p.cntB[pl] ||
        sb % 2 != (p.cntB[pr] - p.cntB[pl]) % 2 ||
        min(sr - sl, s.preA[sr]) == sr - sl &&
            sr - sl == min(pr - pl, p.preA[pr]) && sr - sl < pr - pl)
      cout << 0;
    else
      cout << 1;
  }
  cout << '\n';
}
