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 x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int n, m, sum1[100010], sum2[100010], Q, la1[100010], la2[100010];
char s1[100010], s2[100010], ans[100010];
int main() {
  scanf("%s", s1 + 1);
  scanf("%s", s2 + 1);
  n = strlen(s1 + 1);
  m = strlen(s2 + 1);
  Q = read();
  for (int i = 1; i <= n; ++i) {
    sum1[i] = sum1[i - 1] + (s1[i] != 'A');
    if (s1[i] == 'A')
      la1[i] = la1[i - 1] + 1;
    else
      la1[i] = 0;
  }
  for (int i = 1; i <= m; ++i) {
    sum2[i] = sum2[i - 1] + (s2[i] != 'A');
    if (s2[i] == 'A')
      la2[i] = la2[i - 1] + 1;
    else
      la2[i] = 0;
  }
  for (int i = 1; i <= Q; ++i) {
    int l1 = read(), r1 = read(), l2 = read(), r2 = read();
    int cnt1 = sum1[r1] - sum1[l1 - 1], cnt2 = sum2[r2] - sum2[l2 - 1];
    int num1 = min(la1[r1], r1 - l1 + 1), num2 = min(la2[r2], r2 - l2 + 1);
    if (num1 < num2) {
      ans[i] = '0';
      continue;
    }
    if (!cnt1 && cnt2) {
      if (num1 == num2) {
        ans[i] = '0';
        continue;
      }
      cnt1 += 2;
    } else if ((num1 - num2) % 3)
      cnt1 += 2;
    if (cnt1 > cnt2 || (cnt2 - cnt1) & 1) {
      ans[i] = '0';
      continue;
    }
    ans[i] = '1';
  }
  puts(ans + 1);
  return 0;
}
