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 double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
const int MAX = 1000 * 100 + 47;
char AS[MAX];
char BS[MAX];
int A[MAX], B[MAX];
int get(int* A, int l, int r) {
  int sum = A[r];
  if (l) sum -= A[l - 1];
  return sum;
}
int lastZero(int* A, int l, int r) {
  int L = l - 1;
  int R = r + 1;
  while (R - L > 1) {
    int m = (L + R) / 2;
    int s = get(A, m, r);
    if (s == (r - m + 1))
      R = m;
    else
      L = m;
  }
  return L;
}
int solve(int l, int r, int ll, int rr) {
  int a = lastZero(A, l, r);
  int b = lastZero(B, ll, rr);
  int c1 = (r - l + 1) - get(A, l, r);
  int c2 = (rr - ll + 1) - get(B, ll, rr);
  int ac = r - (a + 1) + 1;
  int bc = rr - (b + 1) + 1;
  if (ac < bc) return 0;
  if (ac == bc && c1 == 0 && c2 != 0) return 0;
  ac %= 3;
  bc %= 3;
  if (ac != bc) {
    --ac;
    c1 += 2;
  }
  if (c1 > c2) return 0;
  return (c1 % 2) == (c2 % 2);
}
int main() {
  scanf("%s", AS);
  scanf("%s", BS);
  int n = strlen(AS);
  int m = strlen(BS);
  for (int i = (0); i < (n); i++) A[i] = AS[i] == 'A';
  for (int i = (0); i < (m); i++) B[i] = BS[i] == 'A';
  for (int i = (1); i < (n); i++) A[i] += A[i - 1];
  for (int i = (1); i < (m); i++) B[i] += B[i - 1];
  int q;
  scanf("%d", &q);
  for (int i = (0); i < (q); i++) {
    int l, r, ll, rr;
    scanf("%d%d%d%d", &l, &r, &ll, &rr);
    --l, --r, --ll, --rr;
    printf("%d", solve(l, r, ll, rr));
  }
  printf("\n");
  return 0;
}
