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, t;
const int N = 100000 + 10;
int q, a, b, c, d, sa[N], ta[N], sb[N], tb[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> s >> t >> q;
  for (int i = 0; i < s.length(); i++) {
    sa[i + 1] = sa[i] + (s[i] != 'A');
    sb[i + 1] = (s[i] == 'A') * (sb[i] + 1);
  }
  for (int i = 0; i < t.length(); i++) {
    ta[i + 1] = ta[i] + (t[i] != 'A');
    tb[i + 1] = (t[i] == 'A') * (tb[i] + 1);
  }
  while (q--) {
    cin >> a >> b >> c >> d;
    if (((ta[d] - ta[c - 1]) & 1) != ((sa[b] - sa[a - 1]) & 1)) {
      cout << '0';
    } else if (ta[d] - ta[c - 1] == sa[b] - sa[a - 1] &&
               (min(tb[d], d - c + 1) > min(sb[b], b - a + 1) ||
                (min(sb[b], b - a + 1) - min(tb[d], d - c + 1)) % 3)) {
      cout << '0';
    } else if (ta[d] - ta[c - 1] > sa[b] - sa[a - 1] &&
               sa[b] - sa[a - 1] != 0 &&
               min(tb[d], d - c + 1) > min(sb[b], b - a + 1)) {
      cout << '0';
    } else if (ta[d] - ta[c - 1] > sa[b] - sa[a - 1] &&
               sa[b] - sa[a - 1] == 0 &&
               min(tb[d], d - c + 1) >= min(sb[b], b - a + 1)) {
      cout << '0';
    } else if (ta[d] - ta[c - 1] < sa[b] - sa[a - 1]) {
      cout << '0';
    } else {
      cout << '1';
    }
  }
}
