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 N = 1e7 + 10;
const int md = 1e9 + 7;
const double EPS = 1e-15;
string a, b;
int q;
int l, r, ll, rr;
int cnt[N];
int cnta[N];
int cnt1[N];
int cnt1a[N];
int lastc[N];
int last1c[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> a;
  for (int i = 0; i < int(a.size()); i++) {
    cnt[i + 1] = cnt[i];
    if (a[i] != 'A') cnt[i + 1]++;
    if (a[i] == 'A')
      lastc[i + 1] = lastc[i] + 1;
    else
      lastc[i + 1] = 0;
  }
  cin >> b;
  for (int i = 0; i < int(b.size()); i++) {
    cnt1[i + 1] = cnt1[i];
    if (b[i] != 'A') cnt1[i + 1]++;
    if (b[i] == 'A')
      last1c[i + 1] = last1c[i] + 1;
    else
      last1c[i + 1] = 0;
  }
  cin >> q;
  for (int i = 1; i <= q; i++) {
    cin >> l >> r >> ll >> rr;
    int x = cnt[r] - cnt[l - 1];
    int y = cnt1[rr] - cnt1[ll - 1];
    int first = min(r - l + 1, lastc[r]);
    int s = min(rr - ll + 1, last1c[rr]);
    if (x % 2 == y % 2 && x <= y &&
        (first > s || (first >= s && (x > 0 || x == y))) &&
        (first % 3 == s % 3 || x < y))
      cout << 1;
    else
      cout << 0;
  }
  return 0;
}
