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 T, m, n, sb1[100005], sb2[100005], la1[100005], la2[100005];
char s1[100005], s2[100005];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> s1 + 1 >> s2 + 1 >> T;
  m = strlen(s1 + 1);
  n = strlen(s2 + 1);
  for (int i = 1; i <= m; i++) {
    sb1[i] = sb1[i - 1] + (s1[i] != 'A');
    if (s1[i] == 'A')
      la1[i] = la1[i - 1] + 1;
    else
      la1[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    sb2[i] = sb2[i - 1] + (s2[i] != 'A');
    if (s2[i] == 'A')
      la2[i] = la2[i - 1] + 1;
    else
      la2[i] = 0;
  }
  int l1, r1, l2, r2;
  while (T--) {
    cin >> l1 >> r1 >> l2 >> r2;
    int sum1 = sb1[r1] - sb1[l1 - 1], sum2 = sb2[r2] - sb2[l2 - 1];
    if (sum1 == 0) {
      if (sum2 == 0) {
        if ((r1 - l1 < r2 - l2 || (r1 - l1) % 3 != (r2 - l2) % 3))
          cout << 0;
        else
          cout << 1;
      } else if (sum2 % 2 == 1)
        cout << 0;
      else if (r1 - l1 + 1 <= la2[r2])
        cout << 0;
      else
        cout << 1;
    } else if (la1[r1] < la2[r2])
      cout << 0;
    else if (la1[r1] == la2[r2]) {
      if (sum1 > sum2 || sum1 % 2 != sum2 % 2)
        cout << 0;
      else
        cout << 1;
    } else {
      if (la1[r1] % 3 != la2[r2] % 3) sum1 += 2;
      if (sum1 > sum2 || sum1 % 2 != sum2 % 2)
        cout << 0;
      else
        cout << 1;
    }
  }
}
