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 N = 1e5 + 10;
int n, m, ScntA[N], ScntB[N], TcntA[N], TcntB[N], q, a, b, c, d;
string s, t, res;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> s >> t;
  n = s.size();
  m = t.size();
  for (int i = 0; i < n; i++)
    if (s[i] == 'C') s[i] = 'B';
  for (int i = 0; i < m; i++)
    if (t[i] == 'C') t[i] = 'B';
  for (int i = 1; i <= n; i++) {
    if (s[i - 1] == 'A') ScntA[i] = ScntA[i - 1] + 1;
    ScntB[i] = (s[i - 1] == 'B') + ScntB[i - 1];
  }
  for (int i = 1; i <= m; i++) {
    if (t[i - 1] == 'A') TcntA[i] = TcntA[i - 1] + 1;
    TcntB[i] = (t[i - 1] == 'B') + TcntB[i - 1];
  }
  cin >> q;
  while (q--) {
    cin >> a >> b >> c >> d;
    bool fail = 0;
    int x = ScntB[b] - ScntB[a - 1], y = TcntB[d] - TcntB[c - 1];
    int i = (b - ScntA[b] >= a ? ScntA[b] : b - a + 1);
    int j = (d - TcntA[d] >= c ? TcntA[d] : d - c + 1);
    if (x > y || x % 2 != y % 2) fail = 1;
    if (j > i) fail = 1;
    if (i % 3 != j % 3 && x == y) fail = 1;
    if (x < y && i == j && i == b - a + 1) fail = 1;
    res += (fail ? '0' : '1');
  }
  cout << res;
  return 0;
}
