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 = 1e5 + 10;
int sCountBC[MAXN], tCountBC[MAXN];
inline int sum(int count[MAXN], int a, int b) {
  return count[b + 1] - count[a];
}
int countA(int count[MAXN], int a, int b) {
  if (sum(count, b, b) != 0) return 0;
  int lo = a - 1, hi = b;
  while (hi - lo > 1) {
    int mid = (lo + hi) / 2;
    if (sum(count, mid, b) == 0)
      hi = mid;
    else
      lo = mid;
  }
  return b + 1 - hi;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  string s, t;
  cin >> s >> t;
  for (int i = 0; i < int(int((s).size())); i++)
    sCountBC[i + 1] = sCountBC[i] + (s[i] != 'A');
  for (int i = 0; i < int(int((t).size())); i++)
    tCountBC[i + 1] = tCountBC[i] + (t[i] != 'A');
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a--, b--, c--, d--;
    int sBC = sum(sCountBC, a, b), tBC = sum(tCountBC, c, d),
        sA = countA(sCountBC, a, b), tA = countA(tCountBC, c, d);
    bool ans = tBC >= sBC && sBC % 2 == tBC % 2;
    if (sBC == tBC) {
      ans &= tA <= sA && sA % 3 == tA % 3;
    } else {
      ans &= tA <= sA;
      ans &= !(sA == tA && tBC && !sBC);
    }
    cout << ans;
  }
  cout << '\n';
  return 0;
}
