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 = 1e5 + 10;
char S[N], T[N];
int bs[N], bt[N];
int find(int L, int R, int* b) {
  int l = L - 1, r = R, ret = 0;
  while (l <= r) {
    int m = (l + r) >> 1;
    if (b[R] - b[m] == 0) {
      ret = m + 1;
      r = m - 1;
    } else
      l = m + 1;
  }
  return ret;
}
int main() {
  scanf("%s%s", S + 1, T + 1);
  int lens = strlen(S + 1), lent = strlen(T + 1);
  for (int i = 1; i <= lens; i++) {
    if (S[i] != 'A') bs[i]++;
    bs[i] += bs[i - 1];
  }
  for (int i = 1; i <= lent; i++) {
    if (T[i] != 'A') bt[i]++;
    bt[i] += bt[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int l1, r1, l2, r2;
    scanf("%d%d%d%d", &l1, &r1, &l2, &r2);
    int q1 = find(l1, r1, bs);
    int q2 = find(l2, r2, bt);
    int flg = 0;
    int rq1 = r1 - q1 + 1, rq2 = r2 - q2 + 1;
    if (rq1 >= rq2) {
      int n = (rq1 - rq2) % 3;
      q1 = bs[r1] - bs[l1 - 1];
      q2 = bt[r2] - bt[l2 - 1];
      if (q2 >= q1 && (q2 - q1) % 2 == 0) {
        if (q2 > q1) {
          if (!(q1 == 0 && rq1 == rq2)) flg = 1;
        } else if (q2 == q1) {
          if (n == 0) flg = 1;
        }
      }
    }
    if (flg)
      putchar('1');
    else
      putchar('0');
  }
  return 0;
}
