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;
using namespace std::chrono;
const int mod = int(1e9) + 7;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
const int magic = 3;
void solve();
void dump(const vector<int>& a) {
  for (int i = 0; i < a.size(); i++) {
    cerr << a[i] << ' ';
  }
  cerr << endl;
}
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
const int MX = 100500;
int cntS[MX][2];
int cntT[MX][2];
void upd(string& s) {
  for (auto& c : s) {
    if (c == 'C') {
      c = 'B';
    }
  }
}
void solve() {
  string s, t;
  cin >> s >> t;
  upd(s);
  upd(t);
  int n = int(s.size());
  int m = int(t.size());
  for (int i = 1; i <= n; i++) {
    cntS[i][0] = cntS[i - 1][0];
    cntS[i][1] = cntS[i - 1][1];
    int c = s[i - 1] - 'A';
    cntS[i][c]++;
    if (c) {
      cntS[i][0] = 0;
    }
  }
  for (int i = 1; i <= m; i++) {
    cntT[i][0] = cntT[i - 1][0];
    cntT[i][1] = cntT[i - 1][1];
    int c = t[i - 1] - 'A';
    cntT[i][c]++;
    if (c) {
      cntT[i][0] = 0;
    }
  }
  int q;
  cin >> q;
  string ans;
  while (q--) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    int x = min(cntS[r1][0], r1 - l1 + 1);
    int y = min(cntT[r2][0], r2 - l2 + 1);
    int diff_a = x - y;
    int b_on_s = cntS[r1][1] - cntS[l1 - 1][1];
    int b_on_t = cntT[r2][1] - cntT[l2 - 1][1];
    int diff_b = b_on_t - b_on_s;
    if (diff_a < 0 || diff_b < 0 || ((diff_b & 1) == 1)) {
      ans += '0';
      continue;
    }
    if (diff_b == 0 && ((diff_a % 3) > 0)) {
      ans += '0';
      continue;
    }
    if (diff_b > 0 && diff_a == 0 && x == r1 - l1 + 1) {
      ans += '0';
      continue;
    }
    ans += '1';
  }
  cout << ans << endl;
}
