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 mx = 1e5 + 100;
int n, m, Tlen, Slen, SendA[mx], TendA[mx];
char Ts[mx], str[mx], ans[mx];
int Tsum[mx], Ssum[mx];
int main() {
  scanf("%s%s", str + 1, Ts + 1);
  Tlen = strlen(Ts + 1);
  Slen = strlen(str + 1);
  for (int i = 1; i <= Slen; i++)
    if (str[i] == 'C') str[i] = 'B';
  for (int i = 1; i <= Tlen; i++)
    if (Ts[i] == 'C') Ts[i] = 'B';
  for (int i = 1; i <= Slen; i++) {
    if (str[i] == 'B')
      SendA[i] = 0, Ssum[i] = Ssum[i - 1] + 1;
    else
      SendA[i] = SendA[i - 1] + 1, Ssum[i] = Ssum[i - 1];
  }
  for (int i = 1; i <= Tlen; i++) {
    if (Ts[i] == 'B')
      TendA[i] = 0, Tsum[i] = Tsum[i - 1] + 1;
    else
      TendA[i] = TendA[i - 1] + 1, Tsum[i] = Tsum[i - 1];
  }
  scanf("%d", &n);
  ans[n] = 0;
  int a, b, c, d;
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int Sa = min(b - a + 1, SendA[b]), Ta = min(d - c + 1, TendA[d]);
    int Sb = Ssum[b] - Ssum[a - 1], Tb = Tsum[d] - Tsum[c - 1];
    if (Sb && !Tb)
      ans[i] = '0';
    else if (Sb == Tb && Sb == 0) {
      if (b - a < d - c || (b - a - d + c) % 3)
        ans[i] = '0';
      else
        ans[i] = '1';
    } else if (Sb > Tb)
      ans[i] = '0';
    else if ((Tb - Sb) % 2)
      ans[i] = '0';
    else if (Sb == 0 && Sa == Ta)
      ans[i] = '0';
    else if (Sa < Ta || (Tb == Sb && Sa && (Sa - Ta) % 3))
      ans[i] = '0';
    else
      ans[i] = '1';
  }
  puts(ans);
  return 0;
}
