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 min(int x, int y) { return x < y ? x : y; }
const int N = 1e5 + 10;
char s[N], T[N];
int f1[N], f2[N], g1[N], g2[N];
int main(int argc, char const *argv[]) {
  scanf("%s", s + 1);
  scanf("%s", T + 1);
  for (int i = 1; s[i] >= 'A' && s[i] <= 'C'; i++) {
    f2[i] = f2[i - 1];
    if (s[i] == 'A')
      f1[i] = f1[i - 1] + 1;
    else
      f2[i]++;
  }
  for (int i = 1; T[i] >= 'A' && T[i] <= 'C'; i++) {
    g2[i] = g2[i - 1];
    if (T[i] == 'A')
      g1[i] = g1[i - 1] + 1;
    else
      g2[i]++;
  }
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    int p1 = min(y1 - x1 + 1, f1[y1]), p2 = min(y2 - x2 + 1, g1[y2]);
    int q1 = f2[y1] - f2[x1 - 1], q2 = g2[y2] - g2[x2 - 1];
    if (q1 > q2 || ((q2 - q1) % 2) || p1 < p2)
      printf("0");
    else if ((q1 == q2 && (p1 - p2) % 3) || (!q1 && q2 && p1 == p2))
      printf("0");
    else
      printf("1");
  }
  return 0;
}
