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;
string s, p;
vector<int> a, b, x, y;
int main() {
  cin >> s >> p;
  a.push_back(0);
  b.push_back(0);
  x.push_back(0);
  y.push_back(0);
  for (int i = 0; i < s.size(); i++) {
    a.push_back(a[a.size() - 1]);
    x.push_back(x[x.size() - 1]);
    if (s[i] != 'A') {
      a[a.size() - 1]++;
      x[x.size() - 1] = i + 1;
    }
  }
  for (int i = 0; i < p.size(); i++) {
    b.push_back(b[b.size() - 1]);
    y.push_back(y[y.size() - 1]);
    if (p[i] != 'A') {
      b[b.size() - 1]++;
      y[y.size() - 1] = i + 1;
    }
  }
  int q;
  int r1, r2;
  cin >> q;
  int i, j, k, l;
  int moze;
  for (; q > 0; q--) {
    moze = 1;
    cin >> i >> j >> k >> l;
    if ((b[l] - b[k - 1]) - (a[j] - a[i - 1]) < 0) {
      moze = 0;
    }
    if (((b[l] - b[k - 1]) - (a[j] - a[i - 1])) % 2 != 0) {
      moze = 0;
    }
    r2 = l - y[l];
    r1 = j - x[j];
    if (r1 > j - i + 1) r1 = j - i + 1;
    if (r2 > l - k + 1) r2 = l - k + 1;
    if (r1 < r2) {
      moze = 0;
    } else if ((a[j] - a[i - 1]) == 0 && (b[l] - b[k - 1]) > 0 && r1 == r2)
      moze = 0;
    else {
      if ((r1 - r2) % 3 != 0 && (b[l] - b[k - 1]) - (a[j] - a[i - 1]) == 0) {
        moze = 0;
      }
    }
    cout << moze;
  }
  return 0;
}
