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 lt[(int)2e5];
;
int ls[(int)2e5];
int cs[(int)2e5];
int ct[(int)2e5];
int main() {
  string s, t;
  cin >> s >> t;
  s = '0' + s;
  t = '0' + t;
  int cur = 0;
  for (int i = 1; i < s.size(); i++) {
    if (s[i] == 'A')
      cur++;
    else
      cur = 0;
    ls[i] = cur;
  }
  cur = 0;
  for (int i = 1; i < t.size(); i++) {
    if (t[i] == 'A')
      cur++;
    else
      cur = 0;
    lt[i] = cur;
  }
  for (int i = 1; i < s.size(); i++) {
    if (s[i] != 'A') cs[i]++;
    cs[i] += cs[i - 1];
  }
  for (int i = 1; i < t.size(); i++) {
    if (t[i] != 'A') ct[i]++;
    ct[i] += ct[i - 1];
  }
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int a, b;
    int c, d;
    cin >> a >> b >> c >> d;
    int t1 = min(d - c + 1, lt[d]);
    int s1 = min(b - a + 1, ls[b]);
    int s2 = cs[b] - cs[a - 1];
    int t2 = ct[d] - ct[c - 1];
    if (t2 < s2 || t2 % 2 != s2 % 2) {
      cout << 0;
      continue;
    }
    if (t1 > s1) {
      cout << 0;
      continue;
    }
    if (t2 == s2) {
      if (t1 % 3 == s1 % 3) {
        cout << 1;
        continue;
      } else {
        cout << 0;
        continue;
      }
    }
    if (t1 == s1) {
      if (s2 == 0) {
        cout << 0;
        continue;
      } else {
        cout << 1;
        continue;
      }
    }
    cout << 1;
  }
}
