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 = 100010;
int sa[maxn], sb[maxn], ta[maxn], tb[maxn];
int n, m;
char s[maxn], t[maxn];
int main() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; i++) {
    sa[i] = s[i] == 'A' ? sa[i - 1] + 1 : 0;
    sb[i] = sb[i - 1] + (s[i] == 'B' || s[i] == 'C');
  }
  for (int i = 1; i <= m; i++) {
    ta[i] = t[i] == 'A' ? ta[i - 1] + 1 : 0;
    tb[i] = tb[i - 1] + (t[i] == 'B' || t[i] == 'C');
  }
  int q, l, r, Sa, Sb, Ta, Tb, lenS, lenT;
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d", &l, &r);
    Sa = min(sa[r], r - l + 1);
    Sb = sb[r] - sb[l - 1];
    lenS = r - l + 1;
    scanf("%d%d", &l, &r);
    Ta = min(ta[r], r - l + 1);
    Tb = tb[r] - tb[l - 1];
    lenT = r - l + 1;
    if (Sb > Tb || (Sb & 1) != (Tb & 1) || Sa < Ta)
      putchar('0');
    else if (Ta < Sa && Tb > Sb ||
             (Sa - Ta) % 3 == 0 && (Ta != lenS || Ta == lenT))
      putchar('1');
    else
      putchar('0');
  }
}
