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;
string s[2];
int cnt[2][N], fa[2][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int x = 0, qwerty = 2; x < qwerty; x++) cin >> s[x];
  for (int z = 0, qwerty = 2; z < qwerty; z++)
    for (int x = 0, qwerty = ((int)(s[z]).size()); x < qwerty; x++) {
      cnt[z][x] = s[z][x] != 'A';
      if (x) cnt[z][x] += cnt[z][x - 1];
      if (s[z][x] == 'A')
        fa[z][x] = 1 + (x ? fa[z][x - 1] : 0);
      else
        fa[z][x] = 0;
    }
  int q;
  cin >> q;
  for (int x = 0, qwerty = q; x < qwerty; x++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a--;
    b--;
    c--;
    d--;
    int o1 = cnt[0][b] - (a ? cnt[0][a - 1] : 0);
    int o2 = cnt[1][d] - (c ? cnt[1][c - 1] : 0);
    bool can = o2 >= o1 && (o2 - o1) % 2 == 0;
    int a0 = min(fa[0][b], b - a + 1), a1 = min(fa[1][d], d - c + 1);
    if (o1 == o2) {
      if (a1 % 3 != a0 % 3 || a1 > a0) can = 0;
    } else {
      if (a1 > a0) can = 0;
      if (o1 == 0 && o2 && a1 == a0) can = 0;
    }
    cout << can;
  }
  cout << "\n";
}
