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;
const long long INF = 0x7f7f7f7f7f7f;
template <typename _T>
void read(_T &x) {
  _T f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
char s[100005], t[100005];
long long n, lens, lent;
long long ans[100005];
long long sum1[100005], sum2[100005];
long long sub1[100005], sub2[100005];
signed main() {
  scanf("%s%s", s + 1, t + 1);
  lens = (long long)strlen(s + 1);
  lent = (long long)strlen(t + 1);
  read(n);
  for (long long i = 1; i <= lens; i++) {
    sum1[i] = sum1[i - 1] + (s[i] != 'A');
    if (s[i] == 'A') sub1[i] = sub1[i - 1] + 1;
  }
  for (long long i = 1; i <= lent; i++) {
    sum2[i] = sum2[i - 1] + (t[i] != 'A');
    if (t[i] == 'A') sub2[i] = sub2[i - 1] + 1;
  }
  for (long long i = 1; i <= n; i++) {
    long long a, b, c, d;
    ans[i] = 1;
    read(a);
    read(b);
    read(c);
    read(d);
    long long tmp1 = sum2[d] - sum2[c - 1], tmp2 = sum1[b] - sum1[a - 1];
    long long nums = min(b - a + 1, sub1[b]), numt = min(d - c + 1, sub2[d]);
    if (tmp1 < tmp2 || numt > nums || ((tmp1 - tmp2) % 2 == 1)) ans[i] = 0;
    long long tt = (nums - numt) % 3 > 0;
    if (tmp1 < tmp2 + tt * 2) ans[i] = 0;
    if (sum1[b] == sum1[a - 1] && tmp1 != tmp2 && nums == numt) ans[i] = 0;
  }
  for (long long i = 1; i <= n; i++) printf("%d", ans[i]);
  return 0;
}
