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;
string a, b;
int nxa[100010], nxb[100010], cta[100010], ctb[100010];
int n;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b;
  cin >> n;
  for (int i = 0; i < a.size(); i++) {
    if (i) nxa[i] += nxa[i - 1];
    if (a[i] != 'A')
      nxa[i]++, cta[i] = 0;
    else
      cta[i] = i ? cta[i - 1] + 1 : 1;
  }
  for (int i = 0; i < b.size(); i++) {
    if (i) nxb[i] += nxb[i - 1];
    if (b[i] != 'A')
      nxb[i]++, ctb[i] = 0;
    else
      ctb[i] = i ? ctb[i - 1] + 1 : 1;
  }
  while (n--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a--, b--, c--, d--;
    int ta = a ? nxa[b] - nxa[a - 1] : nxa[b];
    int tb = c ? nxb[d] - nxb[c - 1] : nxb[d];
    int laa = min(b - a + 1, cta[b]);
    int lab = min(d - c + 1, ctb[d]);
    if (ta > tb)
      cout << 0;
    else if ((tb - ta) & 1)
      cout << 0;
    else if (laa < lab)
      cout << 0;
    else if (laa == lab && laa == b - a + 1 && lab != d - c + 1)
      cout << 0;
    else if (laa > lab && (laa - lab) % 3 && ta + 2 > tb)
      cout << 0;
    else
      cout << 1;
  }
  return 0;
}
