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;
string S, T;
int Q;
int psum[2][100010], la[2][100010];
int calc(int t, int l, int r) { return psum[t][r] - (l ? psum[t][l - 1] : 0); }
char ans[100010];
int main() {
  cin >> S >> T;
  for (int i = 0; i < S.size(); i++)
    if (S[i] == 'C') S[i] = 'B';
  for (int i = 0; i < T.size(); i++)
    if (T[i] == 'C') T[i] = 'B';
  for (int i = 0; i < S.size(); i++) {
    psum[0][i] = S[i] == 'B';
    if (i) psum[0][i] += psum[0][i - 1];
  }
  for (int i = 0; i < T.size(); i++) {
    psum[1][i] = T[i] == 'B';
    if (i) psum[1][i] += psum[1][i - 1];
  }
  for (int i = 0; i < S.size(); i++) {
    la[0][i] = -1;
    if (i) la[0][i] = la[0][i - 1];
    if (S[i] == 'B') la[0][i] = i;
  }
  for (int i = 0; i < T.size(); i++) {
    la[1][i] = -1;
    if (i) la[1][i] = la[1][i - 1];
    if (T[i] == 'B') la[1][i] = i;
  }
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    a--;
    b--;
    c--;
    d--;
    int t1 = calc(0, a, b);
    int t2 = calc(1, c, d);
    if (t1 > t2 || t1 % 2 != t2 % 2) {
      ans[i] = '0';
      continue;
    }
    int s1 = min(b - a + 1, b - la[0][b]);
    int s2 = min(d - c + 1, d - la[1][d]);
    if (!t1 && t2 && s1 == s2) {
      ans[i] = '0';
      continue;
    }
    if (s1 >= s2 && s1 % 3 == s2 % 3) {
      ans[i] = '1';
    } else if (t1 < t2 && s1 > s2) {
      ans[i] = '1';
    } else
      ans[i] = '0';
  }
  for (int i = 0; i < Q; i++) {
    printf("%c", ans[i]);
  }
}
