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 = 1e5;
void read(int &x) {
  char ch;
  bool flag = 0;
  for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1);
       ch = getchar())
    ;
  for (x = 0; isdigit(ch); x = (x << 1) + (x << 3) + ch - 48, ch = getchar())
    ;
  x *= 1 - 2 * flag;
}
void print(int x) {
  if (x > 9) print(x / 10);
  putchar(x % 10 + 48);
}
void write(int x) {
  if (x < 0) putchar('-');
  print(((x) < 0 ? (-(x)) : (x)));
}
char ch[N + 5];
int s1[N + 5], s2[N + 5], len, q, a, b, c, d;
int l1[N + 5], l2[N + 5];
void work() {
  scanf("%s", ch + 1);
  len = strlen(ch + 1);
  for (int i = 1; i <= len; i++)
    s1[i] = s1[i - 1] + (ch[i] != 'A'),
    l1[i] = (ch[i] == 'A' ? l1[i - 1] + 1 : 0);
  scanf("%s", ch + 1);
  len = strlen(ch + 1);
  for (int i = 1; i <= len; i++)
    s2[i] = s2[i - 1] + (ch[i] != 'A'),
    l2[i] = (ch[i] == 'A' ? l2[i - 1] + 1 : 0);
  read(q);
  while (q--) {
    read(a), read(b), read(c), read(d);
    int ss = s1[b] - s1[a - 1], st = s2[d] - s2[c - 1],
        ls = ((b - a + 1) < (l1[b]) ? (b - a + 1) : (l1[b])),
        lt = ((d - c + 1) < (l2[d]) ? (d - c + 1) : (l2[d]));
    if (ss > st) {
      write(0);
      continue;
    }
    if (ss == st) {
      if (ls < lt) {
        write(0);
        continue;
      }
      if ((ls - lt) % 3)
        write(0);
      else
        write(1);
      continue;
    }
    if (ss == 0) {
      if (st % 2) {
        write(0);
        continue;
      }
      if (ls <= lt) {
        write(0);
        continue;
      }
      if ((ls - lt) % 3 > st)
        write(0);
      else
        write(1);
      continue;
    }
    if (ss < st) {
      if ((st - ss) % 2) {
        write(0);
        continue;
      }
      if (ls < lt) {
        write(0);
        continue;
      }
      if ((ls - lt) % 3 > st - ss)
        write(0);
      else
        write(1);
      continue;
    }
  }
}
int main() {
  work();
  return 0;
}
