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;
template <typename T>
inline T read(T& x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
int n, m, Q, i, l1, r1, l2, r2, cnt, sum1[N], sum2[N], a1[N], a2[N];
char s[N], t[N], ans[N];
int main() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  for (i = 1; i <= n; i++) {
    sum1[i] = sum1[i - 1] + (s[i] != 'A');
    if (s[i] == 'A')
      a1[i] = a1[i - 1] + 1;
    else
      a1[i] = 0;
  }
  for (i = 1; i <= m; i++) {
    sum2[i] = sum2[i - 1] + (t[i] != 'A');
    if (t[i] == 'A')
      a2[i] = a2[i - 1] + 1;
    else
      a2[i] = 0;
  }
  for (read(Q); Q--; cnt++) {
    read(l1), read(r1), read(l2), read(r2);
    int suffa1 = min(a1[r1], r1 - l1 + 1), suffa2 = min(a2[r2], r2 - l2 + 1);
    int cntb1 = sum1[r1] - sum1[l1 - 1], cntb2 = sum2[r2] - sum2[l2 - 1];
    if (cntb1 > cntb2)
      ans[cnt] = '0';
    else if (cntb1 == cntb2) {
      if (suffa1 >= suffa2 && (suffa1 - suffa2) % 3 == 0)
        ans[cnt] = '1';
      else
        ans[cnt] = '0';
    } else {
      if ((cntb2 - cntb1) & 1)
        ans[cnt] = '0';
      else if (suffa2 > suffa1)
        ans[cnt] = '0';
      else if (suffa2 == suffa1 && !cntb1)
        ans[cnt] = '0';
      else
        ans[cnt] = '1';
    }
  }
  puts(ans);
  return 0;
}
