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 maxn = 100500;
int ps[maxn][4];
int pt[maxn][4];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  ps[0][0] = ps[0][1] = ps[0][2] = ps[0][3] = pt[0][0] = pt[0][1] = pt[0][2] =
      ps[0][3];
  for (int i = 0; i < n; i++) {
    ps[i + 1][0] = ps[i][0];
    ps[i + 1][1] = ps[i][1];
    ps[i + 1][2] = ps[i][2];
    ps[i + 1][s[i] - 'A']++;
    ps[i + 1][3] = (s[i] == 'A') ? (ps[i][3] + 1) : 0;
  }
  for (int i = 0; i < m; i++) {
    pt[i + 1][0] = pt[i][0];
    pt[i + 1][1] = pt[i][1];
    pt[i + 1][2] = pt[i][2];
    pt[i + 1][t[i] - 'A']++;
    pt[i + 1][3] = (t[i] == 'A') ? (pt[i][3] + 1) : 0;
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int ss[3];
    int ts[3];
    for (int j = 0; j < 3; j++) {
      ss[j] = ps[b][j] - ps[a - 1][j];
      ts[j] = pt[d][j] - pt[c - 1][j];
    }
    int sa = min(ps[b][3], b - a + 1);
    int ta = min(pt[d][3], d - c + 1);
    if (((ss[1] + ss[2] + ts[1] + ts[2]) & 1) != 0) {
      cout << 0;
      continue;
    }
    if (ss[1] + ss[2] > ts[1] + ts[2] || sa < ta ||
        (sa > ta && ss[1] + ss[2] == ts[1] + ts[2] && (sa - ta) % 3 != 0)) {
      cout << 0;
      continue;
    }
    if (ss[1] + ss[2] == 0 && ts[1] + ts[2] != 0 && sa == ta) {
      cout << 0;
      continue;
    }
    cout << 1;
  }
  return 0;
}
