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 N = 100000 + 10;
char s[N], t[N];
int Q;
int s1[N], s2[N];
int p1[N], p2[N];
int main() {
  scanf("%s %s", s + 1, t + 1);
  int sLen = strlen(s + 1);
  int tLen = strlen(t + 1);
  for (int i = 1; i <= sLen; i++) {
    s1[i] = s1[i - 1] + (s[i] == 'B' || s[i] == 'C');
    if (s[i] == 'B' || s[i] == 'C') {
      p1[i] = i;
    } else {
      p1[i] = p1[i - 1];
    }
  }
  for (int i = 1; i <= tLen; i++) {
    s2[i] = s2[i - 1] + (t[i] == 'B' || t[i] == 'C');
    if (t[i] == 'B' || t[i] == 'C') {
      p2[i] = i;
    } else {
      p2[i] = p2[i - 1];
    }
  }
  scanf("%d", &Q);
  while (Q--) {
    int x, y, l, r;
    scanf("%d %d %d %d", &x, &y, &l, &r);
    int B1 = s1[y] - s1[x - 1];
    int B2 = s2[r] - s2[l - 1];
    int A1 = min(y - p1[y], y - x + 1);
    int A2 = min(r - p2[r], r - l + 1);
    bool ok = 1;
    if (A1 < A2) {
      ok = 0;
    } else if ((A1 - A2) % 3) {
      B1 += 2;
    } else {
      if (A1 != A2) {
        B1 += 2;
        if (((B2 - B1) % 2 == 0 && B2 - B1 >= 0)) {
          printf("1");
          continue;
        }
        B1 -= 2;
      }
    }
    if ((B2 - B1) < 0) ok = 0;
    if (B2 > 0 && B1 == 0) ok = 0;
    if ((B2 - B1) % 2) ok = 0;
    printf("%d", ok ? 1 : 0);
  }
  printf("\n");
}
