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 ans, s[2];
int q, numa[2][100010], sumb[2][100010];
char solve(int numa1, int sumb1, int numa2, int sumb2) {
  if (numa1 < numa2) return '0';
  if (!sumb1 && sumb2) {
    if (numa1 == numa2) return '0';
    sumb1 += 2;
  } else if ((numa1 - numa2) % 3 != 0) {
    sumb1 += 2;
  }
  if (sumb1 > sumb2 || (sumb2 - sumb1) % 2 == 1) return '0';
  return '1';
}
int main() {
  ios::sync_with_stdio(false);
  cin >> s[0] >> s[1];
  for (int i = 0; i < 2; i++) {
    for (int j = 1; j <= s[i].length(); j++) {
      sumb[i][j] = sumb[i][j - 1] + (s[i][j - 1] != 'A');
      numa[i][j] = (s[i][j - 1] == 'A') ? numa[i][j - 1] + 1 : 0;
    }
  }
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int numa1 = min(b - a + 1, numa[0][b]);
    int numa2 = min(d - c + 1, numa[1][d]);
    int sumb1 = sumb[0][b] - sumb[0][a - 1];
    int sumb2 = sumb[1][d] - sumb[1][c - 1];
    ans += solve(numa1, sumb1, numa2, sumb2);
  }
  cout << ans << endl;
  return 0;
}
