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 MOD = 1000000007;
void yes() { cout << "1"; }
void no() { cout << "0"; }
struct Sync_stdio {
  Sync_stdio() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
  }
} _sync_stdio;
struct FAIL {
  FAIL() {
    cout << "CHANGE!!!"
         << "\n";
  }
};
struct Z {
  vector<int> cnt_b;
  vector<int> last_a;
};
Z prepare(const string &s) {
  Z res;
  res.cnt_b.resize(s.size() + 1);
  for (int i = (0); i < (s.size()); ++i) {
    res.cnt_b[i + 1] = res.cnt_b[i] + (s[i] != 'A');
  }
  res.last_a.resize(s.size() + 1);
  for (int i = (0); i < (s.size()); ++i) {
    res.last_a[i + 1] = s[i] == 'A' ? res.last_a[i] + 1 : 0;
  }
  return res;
}
int main() {
  string s, t;
  cin >> s >> t;
  Z s_info = prepare(s);
  Z t_info = prepare(t);
  int q;
  cin >> q;
  for (int i = (0); i < (q); ++i) {
    int x1, x2, y1, y2;
    cin >> x1 >> x2 >> y1 >> y2;
    --x1, --x2, --y1, --y2;
    int cnt_b_s = s_info.cnt_b[x2 + 1] - s_info.cnt_b[x1];
    int cnt_b_t = t_info.cnt_b[y2 + 1] - t_info.cnt_b[y1];
    int last_a_s = min(s_info.last_a[x2 + 1], x2 - x1 + 1);
    int last_a_t = min(t_info.last_a[y2 + 1], y2 - y1 + 1);
    if (cnt_b_s == 0 && cnt_b_t > 0 && last_a_s <= last_a_t) {
      no();
      continue;
    }
    if (cnt_b_s > cnt_b_t || (cnt_b_s - cnt_b_t) % 2) {
      no();
      continue;
    }
    if (cnt_b_s != cnt_b_t && last_a_s >= last_a_t) {
      yes();
      continue;
    }
    if (last_a_s < last_a_t || (last_a_s - last_a_t) % 3) {
      no();
      continue;
    }
    yes();
  }
}
