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 int N = int(3e5), mod = int(1e9) + 7;
char second[N], t[N];
int n, m, q;
int a[N], b[N], e[N], o[N];
int main() {
  scanf("%s\n", second + 1);
  n = strlen(second + 1);
  scanf("%s", t + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; i++) {
    e[i] = e[i - 1] + (second[i] != 'A');
    if (second[i] == 'A') a[i] = a[i - 1] + 1;
  }
  for (int i = 1; i <= m; i++) {
    o[i] = o[i - 1] + (t[i] != 'A');
    if (t[i] == 'A') b[i] = b[i - 1] + 1;
  }
  scanf("%d", &q);
  for (int i = 1, x, y, l, r; i <= q; i++) {
    scanf("%d%d%d%d", &x, &y, &l, &r);
    int x1 = e[y] - e[x - 1], x2 = o[r] - o[l - 1];
    if (x1 > x2) {
      printf("0");
      continue;
    }
    if ((x1 & 1) != (x2 & 1)) {
      printf("0");
      continue;
    }
    int a1 = min(a[y], y - x + 1);
    int a2 = min(b[r], r - l + 1);
    if (a2 > a1) {
      printf("0");
      continue;
    }
    if (a1 == a2 && x2 > 0 && a1 == y - x + 1) {
      printf("0");
      continue;
    }
    if (x1 == x2) {
      if (a1 % 3 != a2 % 3) {
        printf("0");
        continue;
      }
    }
    printf("1");
  }
  return 0;
}
