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 int64 = long long;
namespace oct {
inline void ios(bool enable = false, std::string s = "") {
  if (enable) {
    std::string t;
    freopen((t = s + ".in").data(), "r", stdin);
    freopen((t = s + ".out").data(), "w", stdout);
  }
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);
}
}  // namespace oct
const int N = 100010;
char s[N], t[N];
int n, m;
int sa[N], sb[N], ta[N], tb[N];
int main() {
  oct::ios();
  std::cin >> s + 1 >> t + 1;
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; i++) {
    if (s[i] == 'A') sa[i] = sa[i - 1] + 1;
    sb[i] = sb[i - 1] + (s[i] == 'B' || s[i] == 'C');
  }
  for (int i = 1; i <= m; i++) {
    if (t[i] == 'A') ta[i] = ta[i - 1] + 1;
    tb[i] = tb[i - 1] + (t[i] == 'B' || t[i] == 'C');
  }
  int q;
  std::cin >> q;
  int l, r, Sa, Sb, Ta, Tb, lenS, lenT;
  while (q--) {
    std::cin >> l >> r;
    Sa = std::min(r - l + 1, sa[r]);
    Sb = sb[r] - sb[l - 1];
    lenS = r - l + 1;
    std::cin >> l >> r;
    Ta = std::min(r - l + 1, ta[r]);
    Tb = tb[r] - tb[l - 1];
    lenT = r - l + 1;
    if (Tb < Sb || ((Tb & 1) != (Sb & 1)) || Ta > Sa) {
      std::cout << 0;
    } else if ((Ta < Sa && Tb > Sb) ||
               ((Ta != lenS || Ta == lenT) && (Sa - Ta) % 3 == 0)) {
      std::cout << 1;
    } else {
      std::cout << 0;
    }
  }
  std::cout << std::endl;
  return 0;
}
