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 = 2e5;
string s, t;
int a[N], b[N], l1, r1, l2, r2, q, f[N], z[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> s >> t;
  int n = int(s.size());
  int m = int(t.size());
  s = " " + s;
  t = " " + t;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == 'A') f[i] = f[i - 1] + 1;
    b[i] = b[i - 1] + (s[i] == 'B' || s[i] == 'C');
  }
  for (int i = 1; i <= m; ++i) {
    if (t[i] == 'A') z[i] = z[i - 1] + 1;
    a[i] = a[i - 1] + (t[i] == 'B' || t[i] == 'C');
  }
  cin >> q;
  for (int i = 1; i <= q; ++i) {
    cin >> l1 >> r1 >> l2 >> r2;
    int cnt1 = b[r1] - b[l1 - 1];
    int cnt2 = a[r2] - a[l2 - 1];
    if (cnt1 > cnt2) {
      cout << 0;
      continue;
    }
    if ((cnt2 - cnt1) % 2 == 1) {
      cout << 0;
      continue;
    }
    if (!cnt1 && !cnt2) {
      if (r2 - l2 > r1 - l1)
        cout << 0;
      else {
        if ((r1 - l1 - r2 + l2) % 3 == 0)
          cout << 1;
        else
          cout << 0;
      }
      continue;
    }
    if (cnt1 == cnt2) {
      if (z[r2] > f[r1])
        cout << 0;
      else {
        if ((f[r1] - z[r2]) % 3 == 0)
          cout << 1;
        else
          cout << 0;
      }
      continue;
    }
    if (cnt1)
      cout << (f[r1] >= z[r2]);
    else
      cout << ((r1 - l1 + 1) > z[r2]);
  }
  return 0;
}
