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 = 1e5 + 10;
char s[maxN], t[maxN];
int i, a[maxN], b[maxN], n, q, m, x, y, l, r, ans, ss[maxN], st[maxN],
    lass[maxN], last[maxN], s0, t0, s1, t1;
void Y() { printf("1"); }
void N() { printf("0"); }
int main() {
  scanf("%s\n", s + 1), n = strlen(s + 1);
  scanf("%s\n", t + 1), m = strlen(t + 1);
  for (i = 1; i <= n; i++) a[i] = (s[i] == 'A') ? 0 : 1;
  for (i = 1; i <= n; i++)
    ss[i] = ss[i - 1] + a[i], lass[i] = (a[i]) ? 0 : lass[i - 1] + 1;
  for (i = 1; i <= m; i++) b[i] = (t[i] == 'A') ? 0 : 1;
  for (i = 1; i <= m; i++)
    st[i] = st[i - 1] + b[i], last[i] = (b[i]) ? 0 : last[i - 1] + 1;
  for (scanf("%d", &q); q; q--) {
    scanf("%d%d%d%d", &x, &y, &l, &r);
    s0 = ((lass[y]) < (y - x + 1) ? (lass[y]) : (y - x + 1)),
    s1 = ss[y] - ss[x - 1],
    t0 = ((last[r]) < (r - l + 1) ? (last[r]) : (r - l + 1)),
    t1 = st[r] - st[l - 1];
    if (s1 > t1) {
      N();
      continue;
    }
    if (s0 < t0) {
      N();
      continue;
    }
    if ((s1 & 1) != (t1 & 1)) {
      N();
      continue;
    }
    if (s0 == t0) {
      if (t1 && !s1) {
        N();
        continue;
      }
    }
    if (s0 > t0) {
      if (s1 == t1 && (s0 - t0) % 3 != 0) {
        N();
        continue;
      }
    }
    Y();
  }
}
