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 + 20, inf = 2e9 + 20, mod = 1e9 + 7;
const string YN[] = {"NO", "YES"};
string s, t;
int ps[maxn], pt[maxn], prt[maxn], prs[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> t;
  s = 'A' + s;
  t = 'A' + t;
  int y = 0;
  for (int i = 0; i < s.size(); ++i) {
    if (!i) continue;
    ps[i] = ps[i - 1] + (s[i] != 'A');
    if (s[i] != 'A') y = i;
    prs[i] = y;
  }
  y = 0;
  for (int i = 0; i < t.size(); ++i) {
    if (!i) continue;
    pt[i] = pt[i - 1] + (t[i] != 'A');
    if (t[i] != 'A') y = i;
    prt[i] = y;
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int Bs = (ps[b] - ps[a - 1]), Bt = (pt[d] - pt[c - 1]);
    int As = b - max(a - 1, prs[b]), At = d - max(c - 1, prt[d]);
    bool b1 = 1;
    if (Bs > Bt || (Bt - Bs) & 1) b1 = 0;
    if (At > As) b1 = 0;
    if ((Bs == Bt) && ((As - At) % 3)) b1 = 0;
    if (Bs == 0 && Bt > 0 && As == At) b1 = 0;
    cout << b1;
  }
  cout << endl;
  exit(0);
  ;
}
