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;
inline void read(long long &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void read(int &x) {
  char ch;
  bool flag = false;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') flag = true;
  for (x = 0; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  x = flag ? -x : x;
}
inline void write(long long x) {
  static const long long maxlen = 100;
  static char s[maxlen];
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (!x) {
    putchar('0');
    return;
  }
  long long len = 0;
  for (; x; x /= 10) s[len++] = x % 10 + '0';
  for (long long i = len - 1; i >= 0; --i) putchar(s[i]);
}
int n;
int a[1100000];
int b[1100000];
int pre_a[1100000];
int pre_b[1100000];
char s[1100000];
int main() {
  scanf("%s", s);
  int len1 = strlen(s);
  for (int i = 1; i <= len1; i++)
    if (s[i - 1] == 'A') {
      a[i] = a[i - 1];
      pre_a[i] = pre_a[i - 1] + 1;
    } else {
      a[i] = a[i - 1] + 1;
      pre_a[i] = 0;
    }
  scanf("%s", s);
  int len2 = strlen(s);
  for (int i = 1; i <= len2; i++)
    if (s[i - 1] == 'A') {
      b[i] = b[i - 1];
      pre_b[i] = pre_b[i - 1] + 1;
    } else {
      b[i] = b[i - 1] + 1;
      pre_b[i] = 0;
    }
  read(n);
  for (int i = 1; i <= n; i++) {
    int x, y, t, u;
    read(x);
    read(y);
    read(t);
    read(u);
    if (b[u] - b[t - 1] == 0) {
      printf("%d", (a[y] - a[x - 1] == 0) && (u - t <= y - x) &&
                       ((u - t) % 3 == (y - x) % 3));
      continue;
    }
    if (((a[y] - a[x - 1]) % 2 == (b[u] - b[t - 1]) % 2) &&
        (a[y] - a[x - 1] <= b[u] - b[t - 1]))
      if ((pre_b[u] == 0) &&
          ((a[y] - a[x - 1] < b[u] - b[t - 1]) || (pre_a[y] == 0))) {
        printf("1");
        continue;
      } else if ((pre_b[u] == min(pre_a[y], y - x + 1)) &&
                 (pre_a[y] < y - x + 1)) {
        printf("1");
        continue;
      } else if ((min(pre_a[y], y - x + 1) > pre_b[u]) &&
                 (a[y] - a[x - 1] < b[u] - b[t - 1])) {
        printf("1");
        continue;
      } else if ((min(pre_a[y], y - x + 1) > pre_b[u]) &&
                 (min(pre_a[y], y - x + 1) % 3 == pre_b[u] % 3)) {
        printf("1");
        continue;
      }
    printf("0");
  }
  return 0;
}
