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 + 1;
int i, ns, nt, q, a, b, c, d, as[N], bs[N], at[N], bt[N], sa, sb, ta, tb, p;
char s[N], t[N];
int main() {
  scanf("%s%s%d", s + 1, t + 1, &q);
  ns = strlen(s + 1);
  nt = strlen(t + 1);
  for (i = 1; i <= ns; i++)
    if (s[i] == 'A')
      as[i] = as[i - 1] + 1, bs[i] = bs[i - 1];
    else
      as[i] = 0, bs[i] = bs[i - 1] + 1;
  for (i = 1; i <= nt; i++)
    if (t[i] == 'A')
      at[i] = at[i - 1] + 1, bt[i] = bt[i - 1];
    else
      at[i] = 0, bt[i] = bt[i - 1] + 1;
  for (i = 1; i <= q; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    sa = ((as[b]) < (b - a + 1) ? (as[b]) : (b - a + 1));
    sb = bs[b] - bs[a - 1];
    ta = ((at[d]) < (d - c + 1) ? (at[d]) : (d - c + 1));
    tb = bt[d] - bt[c - 1];
    if (sb > tb)
      p = 0;
    else if (sb == tb)
      p = (sa >= ta && (sa - ta) % 3 == 0);
    else if (sb && sb < tb)
      p = (sa >= ta && (tb - sb) % 2 == 0);
    else
      p = (sa > ta && tb % 2 == 0);
    printf("%d", p);
  }
}
