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>
#pragma GCC optimize(2)
using namespace std;
inline void read(long long &x) {
  x = 0;
  long long f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
long long Gcd(long long a, long long b) {
  if (!b) return a;
  return Gcd(b, a % b);
}
inline long long Abs(long long x) {
  if (x < 0) return -x;
  return x;
}
inline double Fabs(double x) {
  if (x > 0) return x;
  return -x;
}
long long PreS[500005], PreT[500005], LasS[500005], LasT[500005];
int main() {
  string S, T;
  cin >> S >> T;
  int lens = S.length(), lent = T.length();
  for (register int i = 1; i <= lens; ++i)
    PreS[i] = PreS[i - 1] + (S[i - 1] == 'B' || S[i - 1] == 'C'),
    LasS[i] = (S[i - 1] == 'B' || S[i - 1] == 'C') ? i : LasS[i - 1];
  for (register int i = 1; i <= lent; ++i)
    PreT[i] = PreT[i - 1] + (T[i - 1] == 'B' || T[i - 1] == 'C'),
    LasT[i] = (T[i - 1] == 'B' || T[i - 1] == 'C') ? i : LasT[i - 1];
  long long Q;
  read(Q);
  while (Q--) {
    long long ls, rs, lt, rt;
    read(ls);
    read(rs);
    read(lt);
    read(rt);
    long long Cha_A = rs - Max(LasS[rs], ls - 1) - rt + Max(LasT[rt], lt - 1);
    long long Cha_B = PreT[rt] - PreT[lt - 1] - PreS[rs] + PreS[ls - 1];
    if (Cha_B % 2 || Cha_B < 0 || Cha_A < 0 ||
        (!Cha_A && Cha_B > 0 && PreS[rs] == PreS[ls - 1]))
      putchar('0');
    else if (!Cha_A || Cha_B > 0 || Cha_A % 3 == 0)
      putchar('1');
    else
      putchar('0');
  }
  return 0;
}
