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;
int len, q, a, b, c, d, i, ss[100001][2], st[100001][2], s0, s1, t0, t1;
char ch[100001];
int pd() {
  if (s1 > t1) return 0;
  if (s0 < t0) return 0;
  if (s1 % 2 != t1 % 2) return 0;
  int x = (t1 - s1) / 2, y = (s0 - t0) % 3;
  if ((s0 > t0) && (x > 0)) return 1;
  if (x < y) return 0;
  if ((!s1) && (t1 > 0) && (!y)) return 0;
  return 1;
}
int main() {
  ss[0][0] = ss[0][1] = st[0][0] = st[0][1] = 0;
  scanf("%s", ch + 1);
  len = strlen(ch + 1);
  for (i = 1; i <= len; i++) {
    ss[i][0] = ss[i - 1][0];
    ss[i][1] = ss[i - 1][1];
    if (ch[i] == 'A')
      ss[i][0]++;
    else {
      ss[i][0] = 0;
      ss[i][1]++;
    }
  }
  scanf("%s", ch + 1);
  len = strlen(ch + 1);
  for (i = 1; i <= len; i++) {
    st[i][0] = st[i - 1][0];
    st[i][1] = st[i - 1][1];
    if (ch[i] == 'A')
      st[i][0]++;
    else {
      st[i][0] = 0;
      st[i][1]++;
    }
  }
  scanf("%d", &q);
  for (i = 1; i <= q; i++) {
    scanf("%d%d%d%d", &a, &b, &c, &d);
    if (ss[b][0] > b - a + 1)
      s0 = b - a + 1;
    else
      s0 = ss[b][0];
    if (st[d][0] > d - c + 1)
      t0 = d - c + 1;
    else
      t0 = st[d][0];
    s1 = ss[b][1] - ss[a - 1][1];
    t1 = st[d][1] - st[c - 1][1];
    printf("%d", pd());
  }
  printf("\n");
}
