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 = 1e5 + 5;
char s[N], t[N];
int a[N], b[N], aA[N], bA[N], m, l, r, L, R;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> (s + 1) >> (t + 1);
  for (int i = 1; s[i]; ++i)
    if (s[i] == 'A')
      a[i] = a[i - 1], aA[i] = aA[i - 1] + 1;
    else
      a[i] = a[i - 1] + 1;
  for (int i = 1; t[i]; ++i)
    if (t[i] == 'A')
      b[i] = b[i - 1], bA[i] = bA[i - 1] + 1;
    else
      b[i] = b[i - 1] + 1;
  for (cin >> m; m--;) {
    cin >> l >> r >> L >> R;
    int x = a[r] - a[l - 1], y = b[R] - b[L - 1];
    if (x > y || (y - x) % 2)
      cout.put('0');
    else {
      int sa = x ? aA[r] : r - l + 1, sb = y ? bA[R] : R - L + 1;
      if (x == y)
        cout.put('0' + (sa >= sb && (sa - sb) % 3 == 0));
      else
        cout.put('0' + (sa > sb || (sa == sb && x)));
    }
  }
  return 0;
}
