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;
char s[100005], t[100005];
int fa[100005], sb[100005], sc[100005], tb[100005], tc[100005], ia[100005];
int res[100005];
int main() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  int n = strlen(s + 1);
  int m = strlen(t + 1);
  for (int i = 1; i <= n; ++i) {
    fa[i] = fa[i - 1] + (s[i] == 'A');
    if (s[i] != 'A') fa[i] = 0;
    sb[i] = sb[i - 1] + (s[i] == 'B');
    sc[i] = sc[i - 1] + (s[i] == 'C');
  }
  for (int i = 1; i <= m; ++i) {
    ia[i] = ia[i - 1] + (t[i] == 'A');
    if (t[i] != 'A') ia[i] = 0;
    tb[i] = tb[i - 1] + (t[i] == 'B');
    tc[i] = tc[i - 1] + (t[i] == 'C');
  }
  int Q;
  scanf("%d", &Q);
  for (int i = 1; i <= Q; ++i) {
    int l, r, ll, rr;
    scanf("%d%d%d%d", &l, &r, &ll, &rr);
    int Sb = sb[r] - sb[l - 1];
    int Sc = sc[r] - sc[l - 1];
    int Tb = tb[rr] - tb[ll - 1];
    int Tc = tc[rr] - tc[ll - 1];
    Sb += Sc;
    Tb += Tc;
    int L = fa[r], R = ia[rr];
    if (L - fa[l - 1] == r - l + 1) L = r - l + 1;
    if (R - ia[ll - 1] == rr - ll + 1) R = rr - ll + 1;
    if (R > L) {
      res[i] = 0;
      continue;
    }
    if ((L - R) % 3 != 0 && L > R) Sb += 2;
    if (L == R && Sb == 0 && Tb) {
      res[i] = 0;
      continue;
    }
    if ((Sb % 2) ^ (Tb % 2) == 1 || Tb < Sb)
      res[i] = 0;
    else
      res[i] = 1;
  }
  for (int i = 1; i <= Q; ++i) printf("%d", res[i]);
  return 0;
}
