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;
long long arr[100005], brr[100005], prea[100005], preb[100005];
char ar[100005], br[100005];
long long n, x, a, b, c, d;
string str;
int main() {
  cin >> ar + 1 >> br + 1;
  x = strlen(ar + 1);
  for (int i = 1; i < x + 1; i++) {
    prea[i] = (ar[i] == 'A') ? 1 + prea[i - 1] : 0;
    arr[i] = (ar[i] != 'A') + arr[i - 1];
  }
  x = strlen(br + 1);
  for (int i = 1; i < x + 1; i++) {
    preb[i] = (br[i] == 'A') ? 1 + preb[i - 1] : 0;
    brr[i] = (br[i] != 'A') + brr[i - 1];
  }
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a >> b >> c >> d;
    long long len1 = min(prea[b], b - a + 1);
    long long len2 = min(preb[d], d - c + 1);
    long long d1 = arr[b] - arr[a - 1];
    long long d2 = brr[d] - brr[c - 1];
    long long flag = 1;
    if (!(d1 % 2 == d2 % 2 && d1 <= d2)) flag = 0;
    if (d1 == d2) {
      if (len1 < len2 || len1 % 3 != len2 % 3) flag = 0;
    }
    if (d1 < d2) {
      if (len1 < len2) flag = 0;
      if (len1 == len2 && d1 == 0) flag = 0;
    }
    str += (flag ? '1' : '0');
  }
  cout << str;
}
