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 + 20;
string S, T;
int Q, s[N], so[N], sz[N], t[N], to[N], tz[N];
int check(int os, int zs, int ot, int zt) {
  if (os == ot) {
    if (zs == zt) return 1;
    if (zs > zt && (zs - zt) % 3 == 0) return 1;
    return 0;
  }
  if (os > ot) return 0;
  if ((ot - os) % 2 != 0) return 0;
  if (os > 0 && zs == zt) return 1;
  if (zs > zt) return 1;
  return 0;
}
int main() {
  cin >> S >> T;
  for (int i = 1; i <= S.size(); i++) {
    if (S[i - 1] == 'A')
      s[i] = 0;
    else
      s[i] = 1;
    so[i] = so[i - 1] + s[i];
    sz[i] = s[i] ? 0 : sz[i - 1] + 1;
  }
  for (int i = 1; i <= T.size(); i++) {
    if (T[i - 1] == 'A')
      t[i] = 0;
    else
      t[i] = 1;
    to[i] = to[i - 1] + t[i];
    tz[i] = t[i] ? 0 : tz[i - 1] + 1;
  }
  cin >> Q;
  for (int i = 0; i < Q; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int os = so[b] - so[a - 1];
    int zs = sz[b] > b - a + 1 ? b - a + 1 : sz[b];
    int ot = to[d] - to[c - 1];
    int zt = tz[d] > d - c + 1 ? d - c + 1 : tz[d];
    cout << check(os, zs, ot, zt);
  }
  cout << endl;
  return 0;
}
