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>
char S[100010];
char T[100010];
int s1[100010];
int t1[100010];
int s0[100010];
int t0[100010];
int main() {
  scanf("%s", S + 1);
  scanf("%s", T + 1);
  int i;
  int lens = strlen(S + 1);
  for (i = 1; i <= lens; i++) {
    if (S[i] == 'B' || S[i] == 'C')
      s1[i] = s1[i - 1] + 1;
    else
      s1[i] = s1[i - 1];
    if (S[i] == 'A')
      s0[i] = s0[i - 1] + 1;
    else
      s0[i] = 0;
  }
  int lent = strlen(T + 1);
  for (i = 1; i <= lent; i++) {
    if (T[i] == 'B' || T[i] == 'C')
      t1[i] = t1[i - 1] + 1;
    else
      t1[i] = t1[i - 1];
    if (T[i] == 'A')
      t0[i] = t0[i - 1] + 1;
    else
      t0[i] = 0;
  }
  int q;
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int ns1 = s1[b] - s1[a - 1];
    int nt1 = t1[d] - t1[c - 1];
    int ns0 = s0[b];
    if (ns0 > b - a + 1) ns0 = b - a + 1;
    int nt0 = t0[d];
    if (nt0 > d - c + 1) nt0 = d - c + 1;
    if (ns1 > nt1 || ns1 % 2 != nt1 % 2 || ns0 < nt0 ||
        (ns1 == nt1 && ns0 % 3 != nt0 % 3) ||
        (ns1 == 0 && nt1 > 0 && ns0 == nt0))
      printf("%d", 0);
    else
      printf("%d", 1);
  }
  return 0;
}
