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 sa[100005], sb[100005];
int ta[100005], tb[100005];
int main() {
  cin >> s;
  int sl = s.length();
  for (int i = 0; i < sl; i++) {
    sb[i] = sb[i - 1] + (s[i] != 'A');
    if (s[i] == 'A') sa[i] = sa[i - 1] + 1;
  }
  cin >> t;
  int tl = t.length();
  for (int i = 0; i < tl; i++) {
    tb[i] = tb[i - 1] + (t[i] != 'A');
    if (t[i] == 'A') ta[i] = ta[i - 1] + 1;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    a--, b--, c--, d--;
    int db = (tb[d] - tb[c - 1]) - (sb[b] - sb[a - 1]);
    int as = min(b - a + 1, sa[b]), at = min(d - c + 1, ta[d]);
    bool f = 1;
    if ((db & 1) || db < 0 || as < at) f = 0;
    if ((as - at) % 3 != 0 && db < 2) f = 0;
    if ((as - at) % 3 == 0 && db < 0) f = 0;
    if (sb[b] == sb[a - 1] && db && as == at) f = 0;
    printf("%d", f);
  }
  return 0;
}
