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")
#pragma GCC target("sse3", "sse2", "sse")
#pragma GCC target("avx", "sse4", "sse4.1", "sse4.2", "ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline", "fast-math", "unroll-loops", \
                     "no-stack-protector")
using namespace std;
template <typename Iter>
ostream& _out(ostream& s, Iter b, Iter e) {
  s << "[";
  for (auto it = b; it != e; it++) s << (it == b ? "" : " ") << *it;
  s << "]";
  return s;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ", " << p.second << ')';
}
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  return in >> p.first >> p.second;
}
template <typename T>
ostream& operator<<(ostream& s, const vector<T>& c) {
  return _out(s, c.begin(), c.end());
}
template <typename T, size_t N>
ostream& operator<<(ostream& s, const array<T, N>& c) {
  return _out(s, c.begin(), c.end());
}
inline char readchar() {
  const int S = 1 << 20;
  static char buf[S], *p = buf, *q = buf;
  if (p == q && (q = (p = buf) + fread(buf, 1, S, stdin)) == buf) return EOF;
  return *p++;
}
inline void input(int& _x) {
  _x = 0;
  int _tmp = 1;
  char _tc = readchar();
  while (_tc < '0' || _tc > '9') _tc = readchar();
  while (_tc >= '0' && _tc <= '9')
    _x = (_x << 1) + (_x << 3) + (_tc - 48), _tc = readchar();
}
inline void output(int _x) {
  char _buff[20];
  int _f = 0;
  if (_x == 0) putchar('0');
  while (_x > 0) {
    _buff[_f++] = _x % 10 + '0';
    _x /= 10;
  }
  for (_f -= 1; _f >= 0; _f--) putchar(_buff[_f]);
  putchar('\n');
}
vector<int> u, v, uu, vv;
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  ;
  string s, t;
  cin >> s >> t;
  u.push_back(0), v.push_back(0);
  uu.push_back(0);
  vv.push_back(0);
  for (auto c : s) u.push_back(c != 'A');
  for (auto c : t) v.push_back(c != 'A');
  for (auto c : s) uu.push_back(c == 'A' ? 1 + uu.back() : 0);
  for (auto c : t) vv.push_back(c == 'A' ? 1 + vv.back() : 0);
  for (int i = 1; i < u.size(); i++) u[i] += u[i - 1];
  for (int i = 1; i < v.size(); i++) v[i] += v[i - 1];
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    if ((v[d] - v[c - 1]) < (u[b] - u[a - 1]))
      putchar('0');
    else if (((v[d] - v[c - 1]) - (u[b] - u[a - 1])) & 1)
      putchar('0');
    else if (min(uu[b], b - a + 1) < min(vv[d], d - c + 1))
      putchar('0');
    else if (u[b] - u[a - 1] == 0 && (v[d] - v[c - 1]) > 0 &&
             min(uu[b], b - a + 1) == min(vv[d], d - c + 1))
      putchar('0');
    else if ((v[d] - v[c - 1]) > (u[b] - u[a - 1]) &&
             min(uu[b], b - a + 1) > min(vv[d], d - c + 1))
      putchar('1');
    else if ((min(uu[b], b - a + 1) - min(vv[d], d - c + 1)) % 3 != 0)
      putchar('0');
    else
      putchar('1');
  }
}
