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 Q, N, M, K, L;
char A[100001], B[100001];
int a1[100001], a2[100001], b1[100001], b2[100001];
int aa1[100001], aa2[100001];
int main() {
  A[0] = B[0] = '0';
  cin >> A + 1 >> B + 1 >> Q;
  N = strlen(A);
  M = strlen(B);
  vector<int> ret;
  for (int i = 1; i < N; i++) {
    a1[i] = a1[i - 1] + (A[i] == 'A');
    b1[i] = b1[i - 1] + (A[i] == 'B' || A[i] == 'C');
    if (A[i] == 'A')
      aa1[i] = aa1[i - 1] + 1;
    else
      aa1[i] = 0;
  }
  for (int i = 1; i < M; i++) {
    a2[i] = a2[i - 1] + (B[i] == 'A');
    b2[i] = b2[i - 1] + (B[i] == 'B' || B[i] == 'C');
    if (B[i] == 'A')
      aa2[i] = aa2[i - 1] + 1;
    else
      aa2[i] = 0;
  }
  for (int q = 0; q < Q; q++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int ca1 = a1[b] - a1[a - 1];
    int cb1 = b1[b] - b1[a - 1];
    int caa1 = aa1[b];
    int ca2 = a2[d] - a2[c - 1];
    int cb2 = b2[d] - b2[c - 1];
    int caa2 = aa2[d];
    if (caa1 > (b - a + 1)) caa1 = aa1[b] - aa1[a - 1];
    if (caa2 > (d - c + 1)) caa2 = aa2[d] - aa2[c - 1];
    if (cb1 <= cb2 && cb1 % 2 == cb2 % 2 && caa1 >= caa2) {
      if (cb1 == 0 && cb2 > 0 && caa1 == caa2)
        ret.push_back(0);
      else if (caa1 % 3 != caa2 % 3 && cb1 + 2 > cb2)
        ret.push_back(0);
      else
        ret.push_back(1);
    } else
      ret.push_back(0);
  }
  for (auto i : ret) cout << i;
  cin >> N;
}
