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>
#pragma GCC optimize "-O3"
using namespace std;
const int MAXN = 230000;
string s, t;
int sm[MAXN];
int pr[MAXN];
pair<int, int> get(int l, int r) {
  int cnt1 = sm[r] - sm[l];
  int cnt0 = r - l;
  if (pr[r] >= l) cnt0 = r - pr[r] - 1;
  return make_pair(cnt1, cnt0);
}
int solve(pair<int, int> x, pair<int, int> y) {
  int a, b, c, d;
  tie(a, b) = x, tie(c, d) = y;
  if ((a + c) % 2 != 0) return 0;
  if (a > c) return 0;
  if (d > b) return 0;
  if (a == c)
    if (b % 3 != d % 3) return 0;
  if (c > a && a == 0 && b == d) {
    return 0;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> s >> t;
  int sh = s.size();
  s += t;
  int n = s.size();
  sm[0] = 0;
  pr[0] = -1;
  for (int i = 0; i < n; ++i) {
    if (s[i] == 'A') {
      sm[i + 1] = sm[i];
      pr[i + 1] = pr[i];
    } else {
      sm[i + 1] = sm[i] + 1;
      pr[i + 1] = i;
    }
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    --l1, --l2;
    l2 += sh, r2 += sh;
    if (solve(get(l1, r1), get(l2, r2)))
      cout << 1;
    else
      cout << 0;
  }
  cout << "\n";
  return 0;
}
