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;
const int maxn = 100500;
int n[2], q;
string s[2];
int pref[2][maxn + 1];
int lastb[2][maxn];
bool query(int l1, int r1, int l2, int r2) {
  int cntb1 = pref[0][r1] - pref[0][l1];
  int cntb2 = pref[1][r2] - pref[1][l2];
  if (cntb1 % 2 != cntb2 % 2) return false;
  if (cntb1 > cntb2) return false;
  int sufa1 = r1 - 1 - max(lastb[0][r1 - 1], l1);
  int sufa2 = r2 - 1 - max(lastb[1][r2 - 1], l2);
  if (sufa2 > sufa1) return false;
  if (cntb1 == cntb2) {
    if (sufa2 % 3 != sufa1 % 3) return false;
    return true;
  } else {
    return true;
  }
}
int main() {
  srand(131);
  ios_base::sync_with_stdio(false);
  cin >> s[0] >> s[1];
  for (int it = 0; it < 2; ++it) {
    n[it] = s[it].size();
    pref[it][0] = 0;
    for (int i = 0; i < n[it]; ++i) {
      if (s[it][i] == 'C') s[it][i] = 'B';
      pref[it][i + 1] = pref[it][i] + (s[it][i] == 'B' ? 1 : 0);
      lastb[it][i] = i ? lastb[it][i - 1] : -1;
      if (s[it][i] == 'B') lastb[it][i] = i;
    }
  }
  cin >> q;
  string ans(q, 'x');
  for (int it = 0; it < q; ++it) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    --l1, --l2;
    ans[it] = query(l1, r1, l2, r2) ? '1' : '0';
  }
  cout << ans << "\n";
  return 0;
}
