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>
#pragma GCC optimize("Ofast")
using namespace std;
template <class T>
T Gcd(T x, T y) {
  return y ? Gcd(y, x % y) : x;
}
template <class T>
T Lcm(T x, T y) {
  return x / Gcd(x, y) * y;
}
template <class T>
void Adj(T &x, T y) {
  if (x >= y) x %= y;
  while (x < 0) x += y;
}
const long long MOD = 1000000007LL;
const int M = 5000005;
const int N = 200005;
int _, l, ll, a[N], aa[N], t1, t2, t3, t4, t5, t6, b[N], bb[N], t7, t8;
char s[N], ss[N];
int main() {
  while (scanf("%s", (s + 1)), !strlen((s + 1)))
    ;
  while (scanf("%s", (ss + 1)), !strlen((ss + 1)))
    ;
  l = strlen(s + 1);
  ll = strlen(ss + 1);
  for (int i = 1; i <= l; i++) {
    a[i] = a[i - 1] + (s[i] == 'B' || s[i] == 'C');
    b[i] = (s[i] == 'A') ? b[i - 1] + 1 : 0;
  }
  for (int i = 1; i <= ll; i++) {
    aa[i] = aa[i - 1] + (ss[i] == 'B' || ss[i] == 'C');
    bb[i] = (ss[i] == 'A') ? bb[i - 1] + 1 : 0;
  }
  for (scanf("%d", &_); _; _--) {
    scanf("%d %d %d %d", &t1, &t2, &t3, &t4);
    t5 = a[t2] - a[t1 - 1];
    t6 = aa[t4] - aa[t3 - 1];
    t7 = t5 ? b[t2] : t2 - t1 + 1;
    t8 = t6 ? bb[t4] : t4 - t3 + 1;
    if (t7 == t8) {
      if (t5 <= t6 && (t6 - t5) % 2 == 0) {
        if (t6 > 0 && t5 > 0)
          putchar('1');
        else if (!t6 && !t5)
          putchar('1');
        else
          putchar('0');
      } else
        putchar('0');
      continue;
    }
    if (t5 == t6) {
      if (t7 >= t8 && (t7 - t8) % 3 == 0)
        putchar('1');
      else
        putchar('0');
    } else if (t5 < t6) {
      if (t7 - t8 >= 0 && (t6 - t5) % 2 == 0)
        putchar('1');
      else
        putchar('0');
    } else
      putchar('0');
  }
  return 0;
}
