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;
char s[100005], t[100005];
int A[2][100005], B[2][100005], NA[2][100005];
int main() {
  scanf("%s%s", s, t);
  int n = strlen(s);
  int m = strlen(t);
  int lastB = 0, lastA = 0, pre_B = -1;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'A') {
      A[0][i + 1] = pre_B;
      B[0][i + 1] = lastB;
      lastA++;
      NA[0][i + 1] = lastA;
    } else {
      A[0][i + 1] = i + 1;
      pre_B = i + 1;
      lastB++;
      B[0][i + 1] = lastB;
      NA[0][i + 1] = lastA;
    }
  }
  lastB = lastA = 0, pre_B = -1;
  for (int i = 0; i < m; i++) {
    if (t[i] == 'A') {
      A[1][i + 1] = pre_B;
      B[1][i + 1] = lastB;
      lastA++;
      NA[1][i + 1] = lastA;
    } else {
      A[1][i + 1] = i + 1;
      pre_B = i + 1;
      lastB++;
      B[1][i + 1] = lastB;
      NA[1][i + 1] = lastA;
    }
  }
  int Q;
  scanf("%d", &Q);
  while (Q--) {
    int a, b, c, d, ans;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int las = b - max(A[0][b], a - 1);
    int lat = d - max(A[1][d], c - 1);
    int lbs = B[0][b] - B[0][a - 1];
    int lbt = B[1][d] - B[1][c - 1];
    if (las < lat) {
      printf("0");
      continue;
    }
    if (lbs == 0 && lbt) {
      if (las == lat) {
        printf("0");
        continue;
      }
      lbs += 2;
    } else if ((las - lat) % 3) {
      lbs += 2;
    }
    if (lbs > lbt || (lbt - lbs) & 1) {
      printf("0");
      continue;
    }
    printf("1");
  }
  return 0;
}
