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;
int ns, nt;
string s, t;
int ScntB[100010], ScntA[100010];
int TcntB[100010], TcntA[100010];
int q;
int sol(int ls, int rs, int lt, int rt) {
  int dbS, dbT, daS, daT;
  dbS = ScntB[rs] - ScntB[ls - 1];
  dbT = TcntB[rt] - TcntB[lt - 1];
  daS = min(rs - ls + 1, ScntA[rs]);
  daT = min(rt - lt + 1, TcntA[rt]);
  if (daS < daT) return 0;
  if ((dbS + dbT) & 1) return 0;
  if (dbS > dbT) return 0;
  if (dbS == dbT) {
    if (daS % 3 == daT % 3) return 1;
    return 0;
  }
  if (dbS == 0 && daS == daT) return 0;
  return 1;
}
int main() {
  cin >> s;
  cin >> t;
  ns = s.length();
  nt = t.length();
  for (int i = 0; i < ns; i++) {
    ScntB[i + 1] = ScntB[i];
    if (s[i] == 'A') {
      if (i > 0 && s[i - 1] == 'A')
        ScntA[i + 1] = ScntA[i] + 1;
      else
        ScntA[i + 1] = 1;
    } else
      ScntB[i + 1] += 1;
  }
  for (int i = 0; i < nt; i++) {
    TcntB[i + 1] = TcntB[i];
    if (t[i] == 'A') {
      if (i > 0 && t[i - 1] == 'A')
        TcntA[i + 1] = TcntA[i] + 1;
      else
        TcntA[i + 1] = 1;
    } else
      TcntB[i + 1] += 1;
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int ls, rs, lt, rt;
    cin >> ls >> rs >> lt >> rt;
    cout << sol(ls, rs, lt, rt);
  }
  return 0;
}
