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;
static inline void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
void gen(const char s[], int a[], int b[]) {
  for (int i = 0; s[i]; ++i) {
    if (s[i] == 'A') {
      b[i + 1] = b[i];
      a[i + 1] = a[i] + 1;
    } else {
      b[i + 1] = b[i] + 1;
    }
  }
}
bool check(int sa, int sb, int ta, int tb) {
  if (sb == tb) return sa >= ta && (sa - ta) % 3 == 0;
  if (sb == 0) return sa > ta && tb % 2 == 0;
  if (tb < sb) return false;
  return sa >= ta && (tb - sb) % 2 == 0;
}
int main() {
  canhazfast();
  static char s[100016], t[100016], ans[100016];
  static int sa[100016], sb[100016];
  static int ta[100016], tb[100016];
  int q;
  cin >> s >> t >> q;
  gen(s, sa, sb);
  gen(t, ta, tb);
  for (int i = 0; i < q; ++i) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    ans[i] = '0' + check(min(sa[b], b - a + 1), sb[b] - sb[a - 1],
                         min(ta[d], d - c + 1), tb[d] - tb[c - 1]);
  }
  cout << ans << '\n';
  return 0;
}
