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 ll = long long;
using namespace std;
const int INF = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  string s;
  string t;
  getline(cin, s);
  getline(cin, t);
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == 'C') {
      s[i] = 'B';
    }
  };
  for (int i = 0; i < t.size(); ++i) {
    if (t[i] == 'C') {
      t[i] = 'B';
    }
  }
  vector<int> sA(s.size(), 0);
  for (int i = 0; i < s.size(); ++i) {
    if (s[i] == 'A') {
      sA[i] = (i == 0 ? 0 : sA[i - 1]) + 1;
    } else {
      sA[i] = 0;
    }
  }
  vector<int> sB(s.size(), s[0] == 'B' ? 1 : 0);
  for (int i = 1; i < s.size(); ++i) {
    sB[i] = sB[i - 1];
    if (s[i] == 'B') {
      ++sB[i];
    }
  }
  vector<int> tA(t.size(), 0);
  for (int i = 0; i < t.size(); ++i) {
    if (t[i] == 'A') {
      tA[i] = (i == 0 ? 0 : tA[i - 1]) + 1;
    } else {
      tA[i] = 0;
    }
  }
  vector<int> tB(t.size(), t[0] == 'B' ? 1 : 0);
  for (int i = 1; i < t.size(); ++i) {
    tB[i] = tB[i - 1];
    if (t[i] == 'B') {
      ++tB[i];
    }
  }
  int q;
  int a, b, c, d;
  for (cin >> q; q > 0; --q) {
    cin >> a >> b >> c >> d;
    --a;
    --b;
    --c;
    --d;
    int l1 = b - a + 1;
    int l2 = d - c + 1;
    int a1 = sA[b];
    if (a1 > l1) {
      a1 = l1;
    }
    int a2 = tA[d];
    if (a2 > l2) {
      a2 = l2;
    }
    int b1 = sB[b] - (a == 0 ? 0 : sB[a - 1]);
    int b2 = tB[d] - (c == 0 ? 0 : tB[c - 1]);
    if (a1 == l1 && b2 > 0) {
      if (a1 > a2) {
        a1 = a2;
      } else {
        --a1;
      }
      b1 += 2;
    }
    if (a1 == l1 || a2 == l2) {
      if (a1 == l1 && a2 == l2 && a1 >= a2 && (a1 - a2) % 3 == 0) {
        cout << '1';
      } else {
        cout << '0';
      }
      continue;
    }
    if (a1 < a2) {
      cout << '0';
      continue;
    }
    a1 -= a2;
    if (a1 % 3 > 0) {
      b1 += 2;
    }
    if (b1 > b2) {
      cout << '0';
      continue;
    }
    if ((b2 - b1) % 2 == 0) {
      cout << '1';
    } else {
      cout << '0';
    }
  }
  return 0;
}
