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 = 302000;
const int mod = 1e9 + 7;
int bs[N], cs[N], bt[N], ct[N];
int as[N], at[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s, t;
  cin >> s >> t;
  int last = -1;
  for (int i = 0; i < s.length(); ++i) {
    bs[i + 1] = (s[i] == 'B') + bs[i];
    cs[i + 1] = (s[i] == 'C') + cs[i];
    if (s[i] != 'A') last = i + 1;
    as[i + 1] = last;
  }
  last = -1;
  for (int i = 0; i < t.length(); ++i) {
    bt[i + 1] = (t[i] == 'B') + bt[i];
    ct[i + 1] = (t[i] == 'C') + ct[i];
    if (t[i] != 'A') last = i + 1;
    at[i + 1] = last;
  }
  int q;
  cin >> q;
  string res;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int x = bs[b] - bs[a - 1] + cs[b] - cs[a - 1];
    int y = bt[d] - bt[c - 1] + ct[d] - ct[c - 1];
    int ax = min(b - a + 1, b - as[b]);
    int ay = min(d - c + 1, d - at[d]);
    int delta = y - x;
    if (ax == b - a + 1 && ax == ay && delta > 0)
      res += '0';
    else if (ax >= ay && (abs(ax % 3) == abs(ay % 3) || delta > 0) && y >= x &&
             abs(y % 2) == abs(x % 2))
      res += '1';
    else
      res += '0';
  }
  cout << res;
  return 0;
}
