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;
int Q, n, m;
struct Str {
  int n, qz[100100], yi[100100];
  char str[100100];
  inline int calc(int u, int v) {
    return qz[v] - qz[u - 1] + (str[v] == 'A') * 2 * (u <= v);
  }
  inline void pre() {
    int i, j;
    n = strlen(str + 1);
    for (i = 1; i <= n; i++) qz[i] = qz[i - 1] + (str[i] != 'A');
    for (i = 1; i <= n; i++) {
      if (str[i] == 'A')
        yi[i] = yi[i - 1] + 1;
      else
        yi[i] = 0;
    }
  }
} A, B;
int main() {
  int i, j, p, q, l, r, cs, ct;
  scanf("%s%s", A.str + 1, B.str + 1);
  A.pre(), B.pre();
  cin >> Q;
  while (Q--) {
    scanf("%d%d%d%d", &p, &q, &l, &r);
    int a = min(A.yi[q], q - p + 1), b = min(B.yi[r], r - l + 1);
    if (a < b) {
      printf("0");
      continue;
    }
    q -= min(a, b), r -= min(a, b);
    if (p > q && l <= r) {
      printf("0");
      continue;
    }
    q -= min(q - p + 1, A.yi[q]) / 3 * 3;
    if (l > r) {
      if ((q - p + 1) % 3 == 0 && A.yi[q] >= q - p + 1)
        printf("1");
      else
        printf("0");
      continue;
    }
    cs = A.calc(p, q);
    ct = B.calc(l, r);
    if (cs > ct || ((cs - ct) & 1))
      printf("0");
    else
      printf("1");
  }
}
