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 int maxn = 1E5 + 10;
int n, m, q, a[maxn], b[maxn], ma[maxn], mb[maxn];
char s[maxn], t[maxn];
int main() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
  scanf("%s", t + 1);
  m = strlen(t + 1);
  cin >> q;
  for (int i = 1; i <= n; i++) {
    a[i] = a[i - 1] + (s[i] != 'A' ? 1 : 0);
    ma[i] = s[i] == 'A' ? ma[i - 1] + 1 : 0;
  }
  for (int i = 1; i <= m; i++) {
    b[i] = b[i - 1] + (t[i] != 'A' ? 1 : 0);
    mb[i] = t[i] == 'A' ? mb[i - 1] + 1 : 0;
  }
  while (q--) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int x = a[r1] - a[l1 - 1];
    int y = b[r2] - b[l2 - 1];
    if (!x && !y) {
      x = r1 - l1 + 1;
      y = r2 - l2 + 1;
      if (x >= y && (x - y) % 3 == 0)
        putchar('1');
      else
        putchar('0');
    } else if (!x && y) {
      int cur = r1 - l1 + 1;
      if (cur - 1 >= mb[r2])
        putchar((y & 1) ? '0' : '1');
      else
        putchar('0');
    } else {
      if ((x & 1) != (y & 1))
        putchar('0');
      else if (mb[r2] > ma[r1])
        putchar('0');
      else if (x > y)
        putchar('0');
      else if (x < y)
        putchar('1');
      else
        putchar((ma[r1] - mb[r2]) % 3 == 0 ? '1' : '0');
    }
  }
  return 0;
}
