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;
using namespace std;
char s[1000005], t[1000005];
int pre[1000005][2], p2[1000005][2];
int jj0[1000005], jj1[1000005];
int main() {
  ios::sync_with_stdio(false);
  cin >> s + 1 >> t + 1;
  for (int i = 1; s[i]; i++) {
    if (s[i] > 'A') {
      pre[i][s[i] - 'A' - 1]++;
      jj0[i] = 0;
    } else
      jj0[i] = jj0[i - 1] + 1;
    pre[i][0] += pre[i - 1][0];
    pre[i][1] += pre[i - 1][1];
  }
  for (int i = 1; t[i]; i++) {
    if (t[i] > 'A') {
      p2[i][t[i] - 'A' - 1]++;
      jj1[i] = 0;
    } else
      jj1[i] = jj1[i - 1] + 1;
    p2[i][0] += p2[i - 1][0];
    p2[i][1] += p2[i - 1][1];
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int l1, l2, r1, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    int c1 = 0, c2 = 0;
    c1 = pre[r1][0] - pre[l1 - 1][0] + pre[r1][1] - pre[l1 - 1][1];
    c2 = p2[r2][0] - p2[l2 - 1][0] + p2[r2][1] - p2[l2 - 1][1];
    if (c1 > c2) {
      cout << 0;
    } else if (c1 == c2 && c1 == 0) {
      int len1 = r1 - l1 + 1;
      int len2 = r2 - l2 + 1;
      if (len1 < len2 || (len1 - len2) % 3)
        cout << 0;
      else
        cout << 1;
    } else if ((c2 - c1) % 2)
      cout << 0;
    else if (jj0[r1] < jj1[r2] || (((jj0[r1] - jj1[r2]) % 3) && (c1 == c2)) ||
             ((c1 == 0) && ((jj0[r1] - jj0[l1 - 1]) <= jj1[r2])))
      cout << 0;
    else
      cout << 1;
  }
  cout << endl;
  return 0;
}
