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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long llinf = (1LL << 62);
const int inf = (1 << 30);
const int nmax = 1e5 + 50;
const long long mod = 1e9 + 7;
using namespace std;
int n, i, j, nr[2][nmax], ls[2][nmax], a, b, c, d, q;
string st, rs;
char s[2][nmax];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  for (j = 0; j < 2; j++) {
    cin >> st;
    n = (int)st.size();
    for (i = 1; i <= n; i++) {
      s[j][i] = st[i - 1];
      if (s[j][i] == 'C') s[j][i] = 'B';
      nr[j][i] = nr[j][i - 1] + (s[j][i] == 'B');
      ls[j][i] = (s[j][i] == 'B' ? i : ls[j][i - 1]);
    }
  }
  cin >> q;
  while (q--) {
    cin >> a >> b >> c >> d;
    if (nr[0][b] - nr[0][a - 1] > nr[1][d] - nr[1][c - 1] ||
        (nr[1][d] - nr[1][c - 1] - nr[0][b] + nr[0][a - 1]) & 1)
      rs += '0';
    else if (nr[1][d] - nr[1][c - 1] == 0 && nr[0][b] - nr[0][a - 1] == 0) {
      if (b - a < d - c || (d - c - b + a) % 3)
        rs += '0';
      else
        rs += '1';
    } else if (nr[0][b] - nr[0][a - 1] == 0) {
      if (b - a + 1 > d - ls[1][d])
        rs += '1';
      else
        rs += '0';
    } else if (nr[0][b] - nr[0][a - 1] == nr[1][d] - nr[1][c - 1]) {
      if (b - ls[0][b] < d - ls[1][d] || (d - ls[1][d] - b + ls[0][b]) % 3)
        rs += '0';
      else
        rs += '1';
    } else {
      if (b - ls[0][b] < d - ls[1][d])
        rs += '0';
      else
        rs += '1';
    }
  }
  cout << rs << endl;
  return 0;
}
