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 w1[100001], w2[100001], r2[100001], r1[100001];
int main() {
  string s, t;
  int q;
  cin >> s >> t;
  scanf("%d", &q);
  for (int i = 1; i <= s.size(); ++i)
    if (s[i - 1] == 'A') {
      r1[i] += r1[i - 1];
      r2[i] += 1 + r2[i - 1];
    } else
      r1[i] += 1 + r1[i - 1];
  for (int i = 1; i <= t.size(); ++i)
    if (t[i - 1] == 'A') {
      w1[i] += w1[i - 1];
      w2[i] += 1 + w2[i - 1];
    } else
      w1[i] += 1 + w1[i - 1];
  for (int i = 1; i <= q; ++i) {
    int x, y, u, v, r = 0, w = 0, j1, j2;
    scanf("%d%d%d%d", &x, &y, &u, &v);
    r = r1[y] - r1[x - 1], w = w1[v] - w1[u - 1];
    j1 = min(r2[y], y - x + 1), j2 = min(w2[v], v - u + 1);
    if (r > w || (w - r) % 2 != 0 || j1 < j2)
      printf("0");
    else if (r == w && ((j1 - j2) % 3 != 0))
      printf("0");
    else if (j1 == j2 && r != w && r == 0)
      printf("0");
    else
      printf("1");
  }
  return 0;
}
