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 N, L, i, j, SA[100005], SB[100005], TA[100005], TB[100005], X, Y, Saa, Sbc,
    Taa, Tbc;
bool B;
char C[100005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> C;
  L = strlen(C);
  for (i = 1; i <= L; i++) {
    SB[i] = SB[i - 1];
    if (C[i - 1] == 'A') {
      SA[i] = SA[i - 1] + 1;
    } else {
      SB[i]++;
    }
  }
  cin >> C;
  L = strlen(C);
  for (i = 1; i <= L; i++) {
    TB[i] = TB[i - 1];
    if (C[i - 1] == 'A') {
      TA[i] = TA[i - 1] + 1;
    } else {
      TB[i]++;
    }
  }
  cin >> N;
  for (i = 0; i < N; i++) {
    cin >> X >> Y;
    Saa = min(Y - X + 1, SA[Y]);
    Sbc = SB[Y] - SB[X - 1];
    cin >> X >> Y;
    Taa = min(Y - X + 1, TA[Y]);
    Tbc = TB[Y] - TB[X - 1];
    B = 0;
    if (Sbc == 0 && Tbc != 0) {
      if (Tbc % 2 == 0 && Saa - Taa >= 1) {
        B = 1;
      }
    } else if (Saa >= Taa) {
      if (Sbc < Tbc) {
        if ((Tbc - Sbc) % 2 == 0) {
          B = 1;
        }
      }
      if (Sbc == Tbc) {
        if ((Saa - Taa) % 3 == 0) {
          B = 1;
        }
      }
    }
    cout << B;
  }
}
