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 INF = 1e9 + 7;
const int MAXN = 1e6 + 7;
const double EPS = 1e-8;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> s >> t;
  int n = s.size(), m = t.size();
  vector<int> ps(n + 1, 0), pt(m + 1, 0);
  vector<int> psa(n + 1, 0), pta(m + 1, 0);
  for (int i = 0; i < n; i++) {
    ps[i + 1] = ps[i];
    psa[i + 1] = psa[i];
    if (s[i] != 'A') {
      psa[i + 1] = 0;
      ps[i + 1]++;
    } else {
      psa[i + 1]++;
    }
  }
  for (int i = 0; i < m; i++) {
    pt[i + 1] = pt[i];
    pta[i + 1] = pta[i];
    if (t[i] != 'A') {
      pt[i + 1]++;
      pta[i + 1] = 0;
    } else {
      pta[i + 1]++;
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    int sum1 = ps[r1] - ps[l1 - 1];
    int sum2 = pt[r2] - pt[l2 - 1];
    if (sum1 > sum2) {
      cout << 0;
      continue;
    }
    if ((sum2 - sum1) % 2) {
      cout << 0;
      continue;
    }
    int cnta1 = min(psa[r1], r1 - l1 + 1);
    int cnta2 = min(pta[r2], r2 - l2 + 1);
    if (cnta2 > cnta1) {
      cout << 0;
      continue;
    }
    if (sum1 == sum2 && (cnta1 - cnta2) % 3) {
      cout << 0;
      continue;
    }
    if (cnta1 == cnta2 && cnta1 == r1 - l1 + 1 && sum2 > sum1) {
      cout << 0;
      continue;
    }
    cout << 1;
  }
}
