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[200000], t[200000];
int pres[200000], pret[200000];
int bcs[200000], bct[200000];
int n, m, q;
int main() {
  scanf("%s", s + 1);
  scanf("%s", t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; ++i) {
    if (s[i] == 'A')
      pres[i] = pres[i - 1] + 1;
    else
      pres[i] = 0;
    bcs[i] = bcs[i - 1] + (s[i] != 'A');
  }
  for (int i = 1; i <= m; ++i) {
    if (t[i] == 'A')
      pret[i] = pret[i - 1] + 1;
    else
      pret[i] = 0;
    bct[i] = bct[i - 1] + (t[i] != 'A');
  }
  scanf("%d", &q);
  int a, b, c, d, e, f, las, lat;
  for (int i = 0; i < q; ++i) {
    scanf("%d %d %d %d", &a, &b, &c, &d);
    e = bcs[b] - bcs[a - 1];
    f = bct[d] - bct[c - 1];
    if (e > f || (f - e) % 2 != 0) {
      printf("0");
      continue;
    }
    las = pres[b];
    if (las > b - a + 1) las = b - a + 1;
    lat = pret[d];
    if (lat > d - c + 1) lat = d - c + 1;
    if (las < lat) {
      printf("0");
      continue;
    }
    if (las > lat && e == f) {
      if ((las - lat) % 3 != 0) {
        printf("0");
        continue;
      }
    }
    if (las == lat && e == 0 && f > 0) {
      printf("0");
      continue;
    }
    printf("1");
  }
  return 0;
}
