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("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
inline bool setmin(T &a, T b) {
  if (a > b) return a = b, 1;
  return 0;
}
template <class T>
inline bool setmax(T &a, T b) {
  if (a < b) return a = b, 1;
  return 0;
}
template <class T>
inline T fast(T a, T b, T mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
template <class T>
istream &operator>>(istream &os, vector<T> &container) {
  for (auto &u : container) os >> u;
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &container) {
  for (auto &u : container) os << u << " ";
  return os;
}
template <class T>
inline T gcd(T a, T b) {
  while (b) swap(a %= b, b);
  return a;
}
const long long INF = 1e9 + 7;
const long long mod = 998244353;
const long long BIG_INF = 1e18 + 7;
const long long N = 1e6 + 7;
const long long T = 1 << 20;
const long double inf = 1e18;
long long n, m, k, q;
string a, b;
vector<int> pref(string &second) {
  vector<int> ret(1);
  for (auto &u : second) ret.push_back(ret.back() + (u != 'A'));
  return ret;
}
vector<int> przed(string &second) {
  vector<int> ret((int)(second).size() + 1);
  ret.back() = (int)(ret).size() - 1;
  for (int i = 1; i < (int)(ret).size(); i++)
    ret[i] = second[i - 1] == 'A' ? ret[i - 1] : i;
  return ret;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> a >> b;
  vector<int> A = pref(a), B = pref(b);
  vector<int> aa = przed(a), bb = przed(b);
  cin >> q;
  for (int i = 0; i < (q); i++) {
    vector<int> p(4);
    cin >> p;
    int ile_A = A[p[1]] - A[p[0] - 1];
    int ile_B = B[p[3]] - B[p[2] - 1];
    int kon_A = min(p[1] - p[0] + 1, p[1] - aa[p[1]]);
    int kon_B = min(p[3] - p[2] + 1, p[3] - bb[p[3]]);
    if (kon_A and !kon_B)
      cout << !(ile_B < ile_A + 2 * (kon_A % 3 != kon_B % 3) or
                ile_A % 2 != ile_B % 2);
    else if (kon_A == p[1] - p[0] + 1 and kon_A == kon_B)
      cout << (ile_B == 0);
    else
      cout << !(kon_B > kon_A or ile_B < ile_A + 2 * (kon_A % 3 != kon_B % 3) or
                ile_A % 2 != ile_B % 2);
  }
  cout << '\n';
}
