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 sa[100060], sb[100060], ta[100060], tb[100060];
char s[100060], t[100060], ara[100070];
int main() {
  int q;
  scanf("%s%s", s + 1, t + 1);
  int ns = strlen(s + 1);
  int nt = strlen(t + 1);
  for (int i = 1; i <= ns; i++) {
    if (s[i] == 'A')
      sa[i] = sa[i - 1] + 1, sb[i] = sb[i - 1];
    else
      sa[i] = 0, sb[i] = sb[i - 1] + 1;
  }
  for (int i = 1; i <= nt; i++) {
    if (t[i] == 'A')
      ta[i] = ta[i - 1] + 1, tb[i] = tb[i - 1];
    else
      ta[i] = 0, tb[i] = tb[i - 1] + 1;
  }
  int i = -1;
  scanf("%d", &q);
  while (q--) {
    i++;
    int x, y, X, Y;
    scanf("%d%d%d%d", &x, &y, &X, &Y);
    int s_trA = min(y - x + 1, sa[y]), t_trA = min(Y - X + 1, ta[Y]);
    int s_b = sb[y] - sb[x - 1], t_b = tb[Y] - tb[X - 1];
    if (s_b == t_b) {
      if (s_trA >= t_trA && (s_trA - t_trA) % 3 == 0)
        ara[i] = '1';
      else
        ara[i] = '0';
    } else if (s_b > 0 && t_b > 0) {
      if (s_b <= t_b && (t_b - s_b) % 2 == 0 && s_trA >= t_trA)
        ara[i] = '1';
      else
        ara[i] = '0';
    } else if (t_b > 0) {
      if (t_b % 2 == 0 && t_trA < s_trA)
        ara[i] = '1';
      else
        ara[i] = '0';
    } else if (s_b > 0) {
      ara[i] = '0';
    } else {
      if (s_trA >= t_trA && (s_trA - t_trA) % 3 == 0)
        ara[i] = '1';
      else
        ara[i] = '0';
    }
  }
  ara[i + 1] = 0;
  puts(ara);
  return 0;
}
