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 = 100005;
const long long MOD = 1e9 + 7;
char s[N], t[N];
int n, m, A1[N], B1[N], A2[N], B2[N];
int main() {
  scanf("%s%s", s, t);
  n = strlen(s);
  m = strlen(t);
  for (int i = 1; i <= n; ++i) {
    A1[i] = s[i - 1] == 'A' ? A1[i - 1] + 1 : 0;
    B1[i] = B1[i - 1] + (s[i - 1] != 'A');
  }
  for (int i = 1; i <= m; ++i) {
    A2[i] = t[i - 1] == 'A' ? A2[i - 1] + 1 : 0;
    B2[i] = B2[i - 1] + (t[i - 1] != 'A');
  }
  int q, a, b, c, d;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    if (int b1 = B1[b] - B1[a - 1], b2 = B2[d] - B2[c - 1],
        a1 = (b1 ? A1[b] : A1[b] - A1[a - 1]),
        a2 = (b2 ? A2[d] : A2[d] - A2[c - 1]);
        (!b1 && !b2 && a1 >= a2 && a1 % 3 == a2 % 3) ||
        (!b1 && b2 && b2 % 2 == 0 && a1 > a2) ||
        ((b1 && b1 <= b2 && b1 % 2 == b2 % 2) &&
         ((a1 >= a2 && a1 % 3 == a2 % 3) || (a1 > a2 && b1 < b2))))
      printf("1");
    else
      printf("0");
  }
  puts("");
  return 0;
}
