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 sBcCount[100000 + 1], tBcCount[100000 + 1];
int sATail[100000 + 1], tATail[100000 + 1];
void preprocess(string& str, int* bcCount, int* aTail) {
  bcCount[0] = aTail[0] = 0;
  for (int i = 0; i < str.length(); i++) {
    bcCount[i + 1] = bcCount[i] + (str[i] != 'A');
    aTail[i + 1] = str[i] == 'A' ? aTail[i] + 1 : 0;
  }
}
int getBcCount(int* bcCount, int from, int to) {
  return bcCount[to] - bcCount[from - 1];
}
int getATail(int* aTail, int from, int to) {
  return min(aTail[to], to - from + 1);
}
int main() {
  string s, t;
  cin >> s >> t;
  preprocess(s, sBcCount, sATail);
  preprocess(t, tBcCount, tATail);
  int q;
  scanf("%d\n", &q);
  for (int i = 0; i < q; i++) {
    int ai, bi, ci, di;
    cin >> ai >> bi >> ci >> di;
    int sbc = getBcCount(sBcCount, ai, bi), tbc = getBcCount(tBcCount, ci, di),
        sat = getATail(sATail, ai, bi), tat = getATail(tATail, ci, di);
    if (sbc > tbc) {
      cout << 0;
    } else if (sbc == tbc) {
      cout << (sat >= tat && (sat - tat) % 3 == 0);
    } else if (sbc == 0) {
      cout << (tbc % 2 == 0 && sat > tat);
    } else {
      cout << ((tbc - sbc) % 2 == 0 && sat >= tat);
    }
  }
  cout << endl;
  return 0;
}
