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;
char str1[100005], str2[100005];
int len1, len2;
int sbc1[100005], sbc2[100005];
int sua1[100005], sua2[100005];
int Q;
int main() {
  scanf("%s%s", str1 + 1, str2 + 1);
  len1 = strlen(str1 + 1), len2 = strlen(str2 + 1);
  for (int i = 1; i <= (len1); ++i)
    sbc1[i] = sbc1[i - 1] + (str1[i] == 'B' || str1[i] == 'C');
  for (int i = 1; i <= (len2); ++i)
    sbc2[i] = sbc2[i - 1] + (str2[i] == 'B' || str2[i] == 'C');
  for (int i = 1; i <= (len1); ++i)
    sua1[i] = str1[i] == 'A' ? sua1[i - 1] + 1 : 0;
  for (int i = 1; i <= (len2); ++i)
    sua2[i] = str2[i] == 'A' ? sua2[i - 1] + 1 : 0;
  scanf("%d", &Q);
  while (Q--) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int sumbc1 = sbc1[b] - sbc1[a - 1], sumbc2 = sbc2[d] - sbc2[c - 1];
    int sufa1 = min(sua1[b], b - a + 1), sufa2 = min(sua2[d], d - c + 1);
    if ((sumbc1 & 1) != (sumbc2 & 1) || sumbc1 > sumbc2 || sufa1 < sufa2) {
      putchar('0');
      continue;
    }
    if (sumbc1 == sumbc2)
      if (sufa1 % 3 == sufa2 % 3)
        putchar('1');
      else
        putchar('0');
    else if (sumbc1 == 0)
      putchar(sufa1 > sufa2 ? '1' : '0');
    else
      putchar('1');
  }
  return 0;
}
