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 = 100002;
string a, b;
int n, m;
int q;
int la[N], push_back[N];
int la2[N], pb2[N];
int main() {
  cin >> a >> b;
  n = ((int)a.size());
  m = ((int)b.size());
  for (int i = (0); (i) < (n); ++(i)) {
    if (i != 0) push_back[i] = push_back[i - 1];
    if (a[i] == 'A') {
      if (i != 0 && a[i - 1] == 'A')
        la[i] = la[i - 1];
      else
        la[i] = i;
    } else
      ++push_back[i];
  }
  for (int i = (0); (i) < (m); ++(i)) {
    if (i != 0) pb2[i] = pb2[i - 1];
    if (b[i] == 'A') {
      if (i != 0 && b[i - 1] == 'A')
        la2[i] = la2[i - 1];
      else
        la2[i] = i;
    } else
      ++pb2[i];
  }
  scanf("%d", &q);
  for (int i = (0); (i) < (q); ++(i)) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    --r1;
    --l1;
    --r2;
    --l2;
    int one1, one2, zero1, zero2;
    if (l1 == 0)
      one1 = push_back[r1];
    else
      one1 = push_back[r1] - push_back[l1 - 1];
    if (l2 == 0)
      one2 = pb2[r2];
    else
      one2 = pb2[r2] - pb2[l2 - 1];
    if (a[r1] != 'A')
      zero1 = 0;
    else
      zero1 = r1 - la[r1] + 1;
    if (zero1 > r1 - l1 + 1) zero1 = r1 - l1 + 1;
    if (b[r2] != 'A')
      zero2 = 0;
    else
      zero2 = r2 - la2[r2] + 1;
    if (zero2 > r2 - l2 + 1) zero2 = r2 - l2 + 1;
    if (zero1 < zero2) {
      printf("0");
      continue;
    }
    if (one1 > one2) {
      printf("0");
      continue;
    }
    if ((one2 - one1) % 2 == 1) {
      printf("0");
      continue;
    }
    if (2 > one2 - one1 && zero1 > zero2 && (zero1 - zero2) % 3 != 0) {
      printf("0");
      continue;
    }
    if (one1 < one2 && one1 == 0 && zero1 == zero2) {
      printf("0");
      continue;
    }
    printf("1");
  }
  cout << endl;
  return 0;
}
