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 sum2[200000], sum1[200000], cnt1[200000], cnt2[200000];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  string st1, st2;
  cin >> st1 >> st2;
  for (long long i = 0; i < st1.size(); i++)
    if (st1[i] == 'A')
      sum1[i + 1] = 0, cnt1[i + 1] = 1 + cnt1[i];
    else
      sum1[i + 1] = 1, cnt1[i + 1] = 0;
  for (long long i = 0; i < st2.size(); i++)
    if (st2[i] == 'A')
      sum2[i + 1] = 0, cnt2[i + 1] = 1 + cnt2[i];
    else
      sum2[i + 1] = 1, cnt2[i + 1] = 0;
  for (long long i = 1; i <= st1.size(); i++) sum1[i] += sum1[i - 1];
  for (long long i = 1; i <= st2.size(); i++) sum2[i] += sum2[i - 1];
  long long n;
  cin >> n;
  while (n--) {
    long long l1, r1;
    long long l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    long long c1 = min(cnt1[r1], r1 - l1 + 1);
    long long c2 = min(cnt2[r2], r2 - l2 + 1);
    if ((sum1[r1] - sum1[l1 - 1]) % 2 != (sum2[r2] - sum2[l2 - 1]) % 2) {
      cout << 0;
    } else if ((sum1[r1] - sum1[l1 - 1]) > (sum2[r2] - sum2[l2 - 1])) {
      cout << 0;
    } else if ((sum1[r1] - sum1[l1 - 1]) == (sum2[r2] - sum2[l2 - 1])) {
      if (c1 >= c2 && c1 % 3 == c2 % 3)
        cout << 1;
      else
        cout << 0;
    } else {
      if (c1 >= c2 && c1 % 3 == c2 % 3 && !(c1 == (r1 - l1 + 1)))
        cout << 1;
      else {
        if (c1 - 1 >= c2)
          cout << 1;
        else
          cout << 0;
      }
    }
  }
}
