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 P = 1e9 + 7;
int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 1e5 + 7;
string s, t;
int pres[N], pret[N], q, a, b, c, d, num1, num2, lens[N], lent[N], w;
bool check(int x, int y) {
  if (x < y) return 0;
  return ((x - y) % 3) == 0;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> s >> t;
  for (int i = (1); i < ((int)s.size() + 1); i++) {
    pres[i] = pres[i - 1] + (s[i - 1] != 'A');
    if (s[i - 1] == 'A')
      lens[i] = lens[i - 1] + 1;
    else
      lens[i] = 0;
  }
  for (int i = (1); i < ((int)t.size() + 1); i++) {
    pret[i] = pret[i - 1] + (t[i - 1] != 'A');
    if (t[i - 1] == 'A')
      lent[i] = lent[i - 1] + 1;
    else
      lent[i] = 0;
  }
  cin >> q;
  for (int cas = (0); cas < (q); cas++) {
    cin >> a >> b >> c >> d;
    num1 = pres[b] - pres[a - 1];
    num2 = pret[d] - pret[c - 1];
    if (num1 > num2 || (num2 - num1) % 2 != 0 ||
        (num1 == num2 && num1 == 0 && !check(b - a + 1, d - c + 1)))
      cout << 0;
    else {
      if (num1 == num2) {
        if (num1 == 0)
          cout << 1;
        else
          cout << check(lens[b], lent[d]);
      } else {
        if (num1 != 0)
          w = lens[b];
        else
          w = b - a + 1;
        if (num1 == 0) w--;
        if (w >= lent[d])
          cout << 1;
        else
          cout << 0;
      }
    }
  }
  return 0;
}
