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;
using ll = long long;
using K = long double;
const int N = 100005;
int n, m, q;
char s[N], t[N];
int a[N], b[N], sa[N], sb[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> (s + 1) >> (t + 1);
  n = strlen(s + 1);
  m = strlen(t + 1);
  for (int i = 1; i <= n; ++i) {
    a[i] = s[i] == 'A' ? (1 + a[i - 1]) : 0;
    sa[i] = sa[i - 1] + (s[i] != 'A');
  }
  for (int i = 1; i <= m; ++i) {
    b[i] = t[i] == 'A' ? (1 + b[i - 1]) : 0;
    sb[i] = sb[i - 1] + (t[i] != 'A');
  }
  string result;
  cin >> q;
  for (int i = 0; i < (q); ++i) {
    int ls, rs, lt, rt;
    cin >> ls >> rs >> lt >> rt;
    int lenS = min(a[rs], rs - ls + 1);
    int lenT = min(b[rt], rt - lt + 1);
    if (lenT > lenS) {
      result += '0';
      continue;
    }
    int cntS =
        sa[rs] - sa[ls - 1] + (lenT < lenS && (lenS - lenT) % 3 > 0 ? 2 : 0);
    int cntT = sb[rt] - sb[lt - 1];
    if (lenS == lenT && cntS == 0 && cntT != 0) {
      result += '0';
      continue;
    }
    if (cntS > cntT || cntS % 2 != cntT % 2) {
      result += '0';
    } else {
      result += '1';
    }
  }
  cout << result << '\n';
  return 0;
}
