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 oo = 0x3f3f3f3f;
const long long ooo = 9223372036854775807ll;
const int _cnt = 1000 * 1000 + 7;
const int _p = 1000 * 1000 * 1000 + 7;
const int N = 101005;
const double PI = acos(-1.0);
const double eps = 1e-9;
int o(int x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
struct ttype {
  int f[N], s[N], a[N], n, t;
  void init(char *st) {
    n = strlen(st + 1), f[0] = s[0] = a[0] = t = 0;
    for (int i = (1); i <= (n); ++i)
      a[i] = (st[i] != 'A'), s[i] = s[i - 1] + a[i];
    for (int i = (1); i <= (n); ++i) {
      if (a[i]) t = i;
      f[i] = t;
    }
  }
  void Q(int l, int r, int &x, int &y) {
    y = r - ((l - 1) > (f[r]) ? (l - 1) : (f[r])), x = s[r] - s[l - 1];
  }
} S, T;
char st[N];
int l, r, t, x, y, x_, y_, m;
int main() {
  scanf("%s", st + 1), S.init(st);
  scanf("%s", st + 1), T.init(st);
  scanf("%d", &m);
  while (m--) {
    scanf("%d%d", &l, &r), S.Q(l, r, x, y);
    scanf("%d%d", &l, &r), T.Q(l, r, x_, y_);
    t = x + ((y - y_) % 3 > 0) * 2;
    printf("%c", '0' + (y >= y_ && (t == x_ || t < x_ && (t || y > y_)) &&
                        ((t)&1) == ((x_)&1)));
  }
  printf("\n");
  return 0;
}
