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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 1e5 + 5;
char A[N], B[N];
int cnta[N], cntb[N];
int preva[N], prevb[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> (A + 1);
  cin >> (B + 1);
  int n = strlen(A + 1);
  int m = strlen(B + 1);
  for (int i = 1; i <= n; i++) {
    cnta[i] = cnta[i - 1] + (A[i] == 'B' || A[i] == 'C');
    preva[i] = preva[i - 1];
    if (A[i] == 'B' || A[i] == 'C') preva[i] = i;
  }
  for (int i = 1; i <= m; i++) {
    cntb[i] = cntb[i - 1] + (B[i] == 'B' || B[i] == 'C');
    prevb[i] = prevb[i - 1];
    if (B[i] == 'B' || B[i] == 'C') prevb[i] = i;
  }
  int q;
  cin >> q;
  while (q--) {
    int la, ra, lb, rb;
    cin >> la >> ra >> lb >> rb;
    int cntA = cnta[ra] - cnta[la - 1];
    int cntB = cntb[rb] - cntb[lb - 1];
    int diff = cntB - cntA;
    int endwale_s = ra - max(preva[ra], la - 1);
    int endwale_t = rb - max(prevb[rb], lb - 1);
    if (diff > 0) {
      if (diff & 1)
        cout << 0;
      else {
        if (endwale_s > endwale_t)
          cout << 1;
        else if (endwale_t == endwale_s && cntA)
          cout << 1;
        else
          cout << 0;
      }
    } else if (diff == 0) {
      if (endwale_t > endwale_s)
        cout << 0;
      else {
        int diff2 = endwale_s - endwale_t;
        if (diff2 % 3 == 0)
          cout << 1;
        else
          cout << 0;
      }
    } else {
      cout << 0;
    }
  }
  return 0;
}
