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 = 1e5 + 5;
int find_no_A(int a, int b, int sa[]) {
  int la = a, ra = b + 1;
  while (la < ra) {
    int mid = (la + ra) >> 1;
    if (sa[b] - sa[mid - 1] == b - mid + 1)
      ra = mid;
    else
      la = mid + 1;
  }
  return b - la + 1;
}
int sa[N], sb[N];
int ta[N], tb[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  string s, t;
  cin >> s >> t;
  int ls = s.length();
  int lt = t.length();
  s = '_' + s;
  t = '_' + t;
  for (int i = 1; i <= ls; i++) {
    if (s[i] == 'A')
      sa[i] = 1;
    else
      sb[i] = 1;
    sa[i] += sa[i - 1];
    sb[i] += sb[i - 1];
  }
  for (int i = 1; i <= lt; i++) {
    if (t[i] == 'A')
      ta[i] = 1;
    else
      tb[i] = 1;
    ta[i] += ta[i - 1];
    tb[i] += tb[i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int nbs = sb[b] - sb[a - 1];
    int nbt = tb[d] - tb[c - 1];
    int nas, nat;
    if (nbs > nbt || ((nbt - nbs) & 1) ||
        (nas = find_no_A(a, b, sa)) < (nat = find_no_A(c, d, ta))) {
      cout << 0;
    } else {
      if (nbt == nbs && (nas - nat) % 3) {
        cout << 0;
      } else if (!nbs && nbt && nas - 1 < nat) {
        cout << 0;
      } else
        cout << 1;
    }
  }
  cout << "\n";
  return 0;
}
