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>
const int N = 222222;
const int M = 100000007;
long long Pw(long long a, long long b, long long p) {
  long long v = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) v = v * a % p;
  return v;
}
using namespace std;
char A[N], B[N];
int n, m, Q, l1, l2, r1, r2, S[N], T[N], LS[N], LT[N];
int main() {
  scanf("%s", A + 1);
  scanf("%s", B + 1);
  n = strlen(A + 1);
  m = strlen(B + 1);
  for (int i = 1; i <= n; i++)
    if (S[i] = S[i - 1], A[i] == 'B' || A[i] == 'C')
      S[i]++, LS[i] = 0;
    else
      LS[i] = LS[i - 1] + 1;
  for (int i = 1; i <= m; i++)
    if (T[i] = T[i - 1], B[i] == 'B' || B[i] == 'C')
      T[i]++, LT[i] = 0;
    else
      LT[i] = LT[i - 1] + 1;
  scanf("%d", &Q);
  for (; Q--;) {
    scanf("%d", &l1);
    scanf("%d", &r1);
    scanf("%d", &l2);
    scanf("%d", &r2);
    if ((S[r1] - S[l1 - 1]) % 2 != (T[r2] - T[l2 - 1]) % 2 ||
        (S[r1] - S[l1 - 1] > T[r2] - T[l2 - 1]))
      putchar('0');
    else {
      if (min(r1 - l1 + 1, LS[r1]) < min(r2 - l2 + 1, LT[r2]))
        putchar('0');
      else {
        if (S[r1] - S[l1 - 1] == T[r2] - T[l2 - 1]) {
          if (min(r1 - l1 + 1, LS[r1]) % 3 != min(r2 - l2 + 1, LT[r2]) % 3)
            putchar('0');
          else
            putchar('1');
        } else {
          if (!(S[r1] - S[l1 - 1])) {
            if (min(r1 - l1 + 1, LS[r1]) == min(r2 - l2 + 1, LT[r2]))
              putchar('0');
            else
              putchar('1');
          } else
            putchar('1');
        }
      }
    }
  }
}
