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 n, m, q, sum1[100005], sum2[100005], f1[100005], f2[100005];
char s[100005], t[100005];
signed main() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (register int i = 1; i <= n; i++) {
    sum1[i] = sum1[i - 1] + (s[i] != 'A');
    f1[i] = (s[i] == 'A') ? (f1[i - 1] + 1) : 0;
  }
  for (register int i = 1; i <= m; i++) {
    sum2[i] = sum2[i - 1] + (t[i] != 'A');
    f2[i] = (t[i] == 'A') ? (f2[i - 1] + 1) : 0;
  }
  scanf("%d", &q);
  while (q--) {
    int l, r, x, y;
    scanf("%d%d%d%d", &l, &r, &x, &y);
    int t1 = sum1[r] - sum1[l - 1], t2 = sum2[y] - sum2[x - 1];
    int g1 = min(f1[r], r - l + 1), g2 = min(f2[y], y - x + 1);
    if (t1 > t2 || (t2 - t1) & 1 || g2 > g1 || (t1 == t2 && (g2 - g1) % 3) ||
        (t1 == 0 && t2 != 0 && g1 == g2))
      putchar('0');
    else
      putchar('1');
  }
}
