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 MAXN = 200000;
const int MOD = 998244353;
const int INF = 1e9 + 10;
char s[MAXN];
char t[MAXN];
int sa[MAXN], sb[MAXN];
int ta[MAXN], tb[MAXN];
void fillab(char *s, int *a, int *b) {
  int n = strlen(s);
  a[0] = b[0] = 0;
  for (int i = 0; i < n; ++i) {
    a[i + 1] = (s[i] == 'A' ? a[i] + 1 : 0);
    b[i + 1] = b[i] + (s[i] == 'B' || s[i] == 'C' ? 1 : 0);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  scanf("%s", s);
  scanf("%s", t);
  fillab(s, sa, sb);
  fillab(t, ta, tb);
  int Q, ai, bi, ci, di;
  scanf("%d", &Q);
  while (Q--) {
    scanf("%d%d%d%d", &ai, &bi, &ci, &di);
    int san = min(sa[bi], bi - ai + 1), tan = min(ta[di], di - ci + 1);
    int sbn = sb[bi] - sb[ai - 1], tbn = tb[di] - tb[ci - 1];
    if (sbn > tbn || (sbn - tbn) % 2 != 0) {
      putchar('0');
    } else if (sbn == tbn) {
      putchar((san >= tan && (san - tan) % 3 == 0) ? '1' : '0');
    } else if (sbn != 0 && sbn < tbn) {
      putchar(san >= tan ? '1' : '0');
    } else if (sbn == 0) {
      putchar(san > tan ? '1' : '0');
    }
  }
  return 0;
}
