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>
const bool DEBUG = false;
using namespace std;
const long long maxn = 2e5 + 5;
const long long INF = 1e18 + 1;
const long long M = 1e9 + 7;
const int lg = 21;
string s, t;
int q, a[maxn], b[maxn], A[maxn], B[maxn];
void pre() {
  a[0] = (s[0] == 'A');
  A[0] = (t[0] == 'A');
  b[0] = (s[0] == 'B' or s[0] == 'C');
  B[0] = (t[0] == 'B' or t[0] == 'C');
  for (int i = 1; i < s.size(); i++) {
    b[i] = b[i - 1];
    if (s[i] == 'A')
      a[i] = a[i - 1] + 1;
    else
      a[i] = 0, b[i]++;
  }
  for (int i = 1; i < t.size(); i++) {
    B[i] = B[i - 1];
    if (t[i] == 'A')
      A[i] = A[i - 1] + 1;
    else
      A[i] = 0, B[i]++;
  }
}
int cnt(int l, int r, int a[maxn]) {
  if (l == 0) return a[r];
  return a[r] - a[l - 1];
}
int solve(int l1, int r1, int l2, int r2) {
  int b1 = cnt(l1, r1, b), b2 = cnt(l2, r2, B);
  int a1 = min(r1 - l1 + 1, a[r1]), a2 = min(r2 - l2 + 1, A[r2]);
  if (a2 > a1 or b1 > b2 or b1 % 2 != b2 % 2) return 0;
  if (b1 == b2 and (a1 - a2) % 3 != 0) return 0;
  if (b1 < b2 and a1 == r1 - l1 + 1 and a1 == a2) return 0;
  return 1;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> s >> t;
  cin >> q;
  pre();
  while (q--) {
    int l1, l2, r1, r2;
    cin >> l1 >> r1 >> l2 >> r2, l1--, l2--, r1--, r2--;
    cout << solve(l1, r1, l2, r2);
  }
  cout << endl;
  return 0;
}
