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 = 112345;
char s[MAXN], t[MAXN];
int psum_s[MAXN], psum_t[MAXN];
int qtda_s[MAXN], qtda_t[MAXN];
int sz_s, sz_t;
int main() {
  scanf(" %s %s", s, t);
  sz_s = strlen(s);
  sz_t = strlen(t);
  for (int i = 0; i < sz_s; i++) {
    if (s[i] == 'C') s[i] = 'B';
    psum_s[i + 1] = psum_s[i] + ((s[i] == 'B') ? 1 : 0);
    qtda_s[i + 1] = (s[i] == 'B') ? 0 : (qtda_s[i] + 1);
  }
  for (int i = 0; i < sz_t; i++) {
    if (t[i] == 'C') t[i] = 'B';
    psum_t[i + 1] = psum_t[i] + ((t[i] == 'B') ? 1 : 0);
    qtda_t[i + 1] = (t[i] == 'B') ? 0 : (qtda_t[i] + 1);
  }
  int Q, a, b, c, d;
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int qtdb_s = psum_s[b] - psum_s[a - 1];
    int qtdb_t = psum_t[d] - psum_t[c - 1];
    int tra_s = min(b - a + 1, qtda_s[b]);
    int tra_t = min(d - c + 1, qtda_t[d]);
    if ((qtdb_t - qtdb_s) % 2 != 0 || qtdb_t < qtdb_s || tra_t > tra_s) {
      printf("0");
    } else if (qtdb_s == qtdb_t) {
      if ((tra_s - tra_t) % 3 == 0) {
        printf("1");
      } else {
        printf("0");
      }
    } else if (qtdb_s > 0) {
      printf("1");
    } else {
      if (tra_s - tra_t == 0) {
        printf("0");
      } else {
        printf("1");
      }
    }
  }
  printf("\n");
  return 0;
}
