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 sbc[N], sras[N];
int tbc[N], tras[N];
int main() {
  string s, t;
  cin >> s >> t;
  for (int i = 0; i < s.size(); i++) {
    sbc[i + 1] = sbc[i] + (s[i] == 'B' or s[i] == 'C');
    if (s[i] == 'A') sras[i + 1] = sras[i] + 1;
  }
  for (int i = 0; i < t.size(); i++) {
    tbc[i + 1] = tbc[i] + (t[i] == 'B' or t[i] == 'C');
    if (t[i] == 'A') tras[i + 1] = tras[i] + 1;
  }
  int q;
  cin >> q;
  while (q--) {
    int i1, j1, i2, j2;
    cin >> i1 >> j1 >> i2 >> j2;
    i1--, i2--;
    int ra1 = min(sras[j1], j1 - i1);
    int ra2 = min(tras[j2], j2 - i2);
    int bc1 = sbc[j1] - sbc[i1];
    int bc2 = tbc[j2] - tbc[i2];
    if (ra2 % 3 != ra1 % 3) bc1 += 2;
    if (ra2 > ra1 or ra2 == ra1 and bc1 == 0 and bc2 > 0)
      cout << 0;
    else if (bc1 > bc2 or bc1 % 2 != bc2 % 2)
      cout << 0;
    else
      cout << 1;
  }
  cout << endl;
}
