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;
int counters[100001];
int countert[100001];
int countersA[100001];
int countertA[100001];
char s[100001];
char t[100001];
char ans[100001];
bool isOk(int sa, int ta, int sbc, int tbc) {
  if (sa < ta) return false;
  if (sa == ta && sbc == 0 && tbc > 0) return false;
  sbc += (sa > ta && (sa - ta) % 3 != 0 ? 2 : 0);
  return sbc <= tbc && (tbc - sbc) % 2 == 0;
}
int main() {
  scanf("%s %s", s, t);
  int lens = strlen(s);
  int lent = strlen(t);
  for (int i = 0; i < lens; i++)
    if (s[i] != 'A') {
      counters[i + 1] = counters[i] + 1;
    } else {
      counters[i + 1] = counters[i];
      countersA[i + 1] = countersA[i] + 1;
    }
  for (int i = 0; i < lent; i++)
    if (t[i] != 'A') {
      countert[i + 1] = countert[i] + 1;
    } else {
      countert[i + 1] = countert[i];
      countertA[i + 1] = countertA[i] + 1;
    }
  int n;
  cin >> n;
  int a, b, c, d;
  for (int i = 0; i < n; i++) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int sa = countersA[b];
    if (sa > b + 1 - a) sa = b + 1 - a;
    int ta = countertA[d];
    if (ta > d + 1 - c) ta = d + 1 - c;
    int sbc = counters[b] - counters[a - 1];
    int tbc = countert[d] - countert[c - 1];
    if (isOk(sa, ta, sbc, tbc))
      ans[i] = '1';
    else
      ans[i] = '0';
  }
  ans[n] = 0;
  puts(ans);
  return 0;
}
