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 N = 100005;
int n, m, q;
char s[N], t[N];
int pre_s[N], pre_t[N];
int suf_s[N], suf_t[N];
int main() {
  scanf("%s", s + 1), n = strlen(s + 1);
  scanf("%s", t + 1), m = strlen(t + 1);
  for (int i = 1; i <= n; i++)
    pre_s[i] = pre_s[i - 1] + (s[i] == 'A' ? 0 : 1),
    suf_s[i] = (s[i] == 'A' ? suf_s[i - 1] + 1 : 0);
  for (int i = 1; i <= m; i++)
    pre_t[i] = pre_t[i - 1] + (t[i] == 'A' ? 0 : 1),
    suf_t[i] = (t[i] == 'A' ? suf_t[i - 1] + 1 : 0);
  scanf("%d", &q);
  while (q--) {
    int ls, rs, lt, rt;
    scanf("%d%d%d%d", &ls, &rs, &lt, &rt);
    int pres = pre_s[rs] - pre_s[ls - 1];
    int sufs = min(suf_s[rs], rs - ls + 1);
    int pret = pre_t[rt] - pre_t[lt - 1];
    int suft = min(suf_t[rt], rt - lt + 1);
    if (pres % 2 == pret % 2 && pres <= pret &&
        ((sufs > suft && pres < pret) ||
         (sufs >= suft && sufs % 3 == suft % 3)))
      if (sufs == suft && pres == 0 && pret > 0)
        putchar('0');
      else
        putchar('1');
    else
      putchar('0');
  }
  return 0;
}
