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;
int a, b, c, d, e, s1[100007] = {0}, s2[100007] = {0}, t1[100007] = {0},
                   t2[100007] = {0};
string m, n;
int main() {
  cin >> m >> n;
  for (int i = 1; i <= m.size(); i++) {
    if (m[i - 1] == 'B' || m[i - 1] == 'C') s1[i] = 1;
    s1[i] += s1[i - 1];
  }
  for (int i = 1; i <= n.size(); i++) {
    if (n[i - 1] == 'B' || n[i - 1] == 'C') t1[i] = 1;
    t1[i] += t1[i - 1];
  }
  for (int i = 1; i <= m.size(); i++) {
    if (m[i - 1] == 'A')
      s2[i] = s2[i - 1] + 1;
    else
      s2[i] = 0;
  }
  for (int i = 1; i <= n.size(); i++)
    if (n[i - 1] == 'A')
      t2[i] = t2[i - 1] + 1;
    else
      t2[i] = 0;
  cin >> a;
  for (int i = 1; i <= a; i++) {
    cin >> b >> c >> d >> e;
    int l = (t1[e] - t1[d - 1] - s1[c] + s1[b - 1]);
    int r1 = s2[c];
    if (r1 > c - b + 1) r1 = c - b + 1;
    int r2 = t2[e];
    if (r2 > e - d + 1) r2 = e - d + 1;
    if (r1 == c - b + 1 && l > 0) r1--;
    if (l >= 0 && l % 2 == 0)
      if (r1 >= r2 && (r1 % 3 == r2 % 3 || l >= 2)) {
        cout << 1;
        continue;
      }
    cout << 0;
  }
}
