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 s, t;
int i, q, a, b, c, d;
int su[100005], dp[100005], su2[100005], dp2[100005];
bool calc(int b, int a, int b2, int a2) {
  if (b > b2) return false;
  if ((b2 - b) % 2) return false;
  if (b == b2) {
    if (a2 > a) return false;
    if ((a - a2) % 3) return false;
    return true;
  }
  if (b) {
    if (a2 > a) return false;
    return true;
  }
  if (!b) {
    if (a2 >= a) return false;
    return true;
  }
}
int main() {
  cin >> s;
  for (i = 0; i < s.size(); i++) {
    if (s[i] == 'B' || s[i] == 'C') {
      if (i == 0)
        su[i] = 1;
      else
        su[i] = su[i - 1] + 1;
    } else
      su[i] = su[i - 1];
    if (s[i] == 'A') {
      if (i == 0)
        dp[i] = 1;
      else
        dp[i] = dp[i - 1] + 1;
    }
  }
  cin >> t;
  for (i = 0; i < t.size(); i++) {
    if (t[i] == 'B' || t[i] == 'C') {
      if (i == 0)
        su2[i] = 1;
      else
        su2[i] = su2[i - 1] + 1;
    } else
      su2[i] = su2[i - 1];
    if (t[i] == 'A') {
      if (i == 0)
        dp2[i] = 1;
      else
        dp2[i] = dp2[i - 1] + 1;
    }
  }
  cin >> q;
  while (q) {
    q--;
    cin >> a >> b >> c >> d;
    a--;
    b--;
    c--;
    d--;
    cout << calc(su[b] - su[a - 1], min(dp[b], b - a + 1), su2[d] - su2[c - 1],
                 min(dp2[d], d - c + 1));
  }
  return 0;
}
