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 iinf = 1e9 + 10;
const long long inf = 1ll << 60;
const long long mod = 1e9 + 7;
void GG() {
  cout << "0\n";
  exit(0);
}
long long mpow(long long a, long long n, long long mo = mod) {
  long long re = 1;
  while (n > 0) {
    if (n & 1) re = re * a % mo;
    a = a * a % mo;
    n >>= 1;
  }
  return re;
}
long long inv(long long b, long long mo = mod) {
  if (b == 1) return b;
  return (mo - mo / b) * inv(mo % b) % mo;
}
const int maxn = 1e5 + 5;
int p[maxn], q[maxn];
int ta[maxn], sa[maxn];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  string second, t;
  cin >> second >> t;
  int n = second.length(), m = t.length();
  for (int i = 0; i < n; ++i) {
    p[i + 1] = second[i] != 'A';
    p[i + 1] += p[i];
    if (second[i] == 'A') {
      sa[i + 1] = sa[i] + 1;
    } else {
      sa[i + 1] = 0;
    }
  }
  for (int i = 0; i < m; ++i) {
    q[i + 1] = t[i] != 'A';
    q[i + 1] += q[i];
    if (t[i] == 'A') {
      ta[i + 1] = ta[i] + 1;
    } else {
      ta[i + 1] = 0;
    };
  }
  int Q;
  cin >> Q;
  while (Q--) {
    int l, r, L, R;
    cin >> l >> r >> L >> R;
    int a = p[r] - p[l - 1], b = q[R] - q[L - 1];
    ;
    if ((a % 2 != b % 2) || a > b)
      cout << 0;
    else {
      int SA = min(sa[r], r - l + 1);
      int TA = min(ta[R], R - L + 1);
      ;
      if (((a == b) && (SA % 3 != TA % 3)) || (SA < TA))
        cout << 0;
      else {
        if (a == 0 && b != 0 && (SA - 1 < TA))
          cout << 0;
        else
          cout << 1;
      }
    }
  }
}
