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 ll = long long;
using ld = long double;
using ull = unsigned long long;
const int MX = 100 * 1000 + 7;
struct Str {
  int ps[MX];
  int lastB[MX];
  int n;
  Str() = default;
  void init(const string& s) {
    memset(ps, 0, sizeof(ps));
    memset(lastB, 0, sizeof(lastB));
    n = s.length();
    for (int i = 1; i <= n; i++) {
      if (s[i - 1] == 'A') {
        lastB[i] = lastB[i - 1];
        ps[i] = ps[i - 1];
      } else {
        lastB[i] = i;
        ps[i] = ps[i - 1] + 1;
      }
    }
  }
  pair<int, int> gt(int l, int r) {
    int ptr = lastB[r];
    if (ptr < l) {
      return {0, r - l + 1};
    } else {
      return {ps[ptr] - ps[l - 1], r - ptr};
    }
  }
};
Str ss, tt;
bool ok(int b1, int a1, int b2, int a2) {
  bool fl = false;
  if (b1 == 0 && b2 != 0) {
    a1--;
    b1 += 2;
    fl = true;
  }
  if (b1 > b2) return false;
  if (a1 < a2) return false;
  if (abs(b1 - b2) % 2 == 1) return false;
  if (!fl && b1 == b2) {
    return (a1 - a2) % 3 == 0;
  } else {
    return true;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> s >> t;
  ss.init(s);
  tt.init(t);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    auto s1 = ss.gt(l1, r1);
    auto s2 = tt.gt(l2, r2);
    if (ok(s1.first, s1.second, s2.first, s2.second)) {
      cout << "1";
    } else {
      cout << "0";
    }
  }
}
