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, t;
int n, f[N], g[N], f1[N], g1[N];
char check(int x, int y, int x1, int y1) {
  if (x > y) return '0';
  if ((y - x) & 1) return '0';
  if (x1 < y1) return '0';
  if (x == y && ((x1 - y1) % 3 != 0)) return '0';
  if (x1 == y1 && x < y && x == 0) return '0';
  return '1';
}
int main() {
  cin >> s >> t;
  for (int i = 0; i < s.size(); ++i) {
    f[i + 1] = f[i];
    if (s[i] != 'A')
      ++f[i + 1], f1[i + 1] = 0;
    else
      f1[i + 1] = f1[i] + 1;
  }
  for (int i = 0; i < t.size(); ++i) {
    g[i + 1] = g[i];
    if (t[i] != 'A')
      ++g[i + 1], g1[i + 1] = 0;
    else
      g1[i + 1] = g1[i] + 1;
  }
  scanf("%d", &n);
  string ret(n, '1');
  for (int i = 0; i < n; ++i) {
    int a, b, c, d;
    scanf("%d%d%d%d", &a, &b, &c, &d);
    int x = f[b] - f[a - 1], x1 = min(f1[b], b - a + 1);
    int y = g[d] - g[c - 1], y1 = min(g1[d], d - c + 1);
    ret[i] = check(x, y, x1, y1);
  }
  cout << ret;
}
