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 MAXN = 100005;
string ax, bx;
int q, pre1[MAXN], pre2[MAXN], pre3[MAXN], pre4[MAXN], a, b, c, d;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> ax >> bx >> q;
  for (int i = 1; i <= ax.size(); ++i) {
    if (ax[i - 1] == 'A') {
      pre1[i] = pre1[i - 1] + 1;
      pre2[i] = pre2[i - 1];
    } else {
      pre1[i] = 0;
      pre2[i] = pre2[i - 1] + 1;
    }
  }
  for (int i = 1; i <= bx.size(); ++i) {
    if (bx[i - 1] == 'A') {
      pre3[i] = pre3[i - 1] + 1;
      pre4[i] = pre4[i - 1];
    } else {
      pre3[i] = 0;
      pre4[i] = pre4[i - 1] + 1;
    }
  }
  for (int i = 0; i < q; ++i) {
    int xa, xb, ya, yb;
    cin >> a >> b >> c >> d;
    xa = min(pre1[b], b - a + 1);
    ya = min(pre3[d], d - c + 1);
    xb = pre2[b] - pre2[a - 1];
    yb = pre4[d] - pre4[c - 1];
    if (xa < ya || xb % 2 != yb % 2) {
      cout << 0;
      continue;
    }
    if (xa == ya && xb == 0 && yb != 0) {
      cout << 0;
      continue;
    }
    if (xa > ya && (xa - ya) % 3 != 0) {
      xb += 2;
    }
    if (xb <= yb && (yb - xb) % 2 == 0)
      cout << 1;
    else
      cout << 0;
  }
  cout << '\n';
}
