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 int read() {
  int sum = 0, p = 1;
  char ch = getchar();
  while (!(('0' <= ch && ch <= '9') || ch == '-')) ch = getchar();
  if (ch == '-') p = -1, ch = getchar();
  while ('0' <= ch && ch <= '9') sum = sum * 10 + ch - 48, ch = getchar();
  return sum * p;
}
const int maxn = 1e5 + 20;
char s[maxn];
int n, m, a[maxn], b[maxn], pa[maxn], pb[maxn];
char p[maxn];
void init() {
  cin >> (s + 1);
  cin >> (p + 1);
  n = strlen(s + 1);
  m = strlen(p + 1);
  for (int i = (1), _end_ = (n); i <= _end_; i++)
    a[i] = a[i - 1] + (s[i] == 'B') + (s[i] == 'C');
  for (int i = (1), _end_ = (m); i <= _end_; i++)
    b[i] = b[i - 1] + (p[i] == 'B') + (p[i] == 'C');
  for (int i = (1), _end_ = (n); i <= _end_; i++)
    pa[i] = s[i] == 'A' ? pa[i - 1] + 1 : 0;
  for (int i = (1), _end_ = (m); i <= _end_; i++)
    pb[i] = p[i] == 'A' ? pb[i - 1] + 1 : 0;
}
inline void doing() {
  int q = read();
  for (int i = (1), _end_ = (q); i <= _end_; i++) {
    int l = read(), r = read();
    int l1 = read(), r1 = read();
    int x = a[r] - a[l - 1], y = b[r1] - b[l1 - 1];
    int Pa = min(pa[r], r - l + 1), Pb = min(pb[r1], r1 - l1 + 1);
    if (!x && y) {
      if (y % 2 == 0 && Pa - 1 >= Pb) {
        putchar('1');
        continue;
      } else {
        putchar('0');
        continue;
      }
    } else if (x <= y && (y - x) % 2 == 0 && Pa >= Pb &&
               (x < y || (Pa - Pb) % 3 == 0))
      putchar('1');
    else
      putchar('0');
  }
}
int main() {
  init();
  doing();
  return 0;
}
