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 INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const int N = 1e5 + 5;
string s[2];
int a[2][N], b[2][N];
pair<int, int> query(int id, int l, int r) {
  pair<int, int> ret;
  if (l)
    ret.first = b[id][r] - b[id][l - 1];
  else
    ret.first = b[id][r];
  ret.second = min(a[id][r], r - l + 1);
  return ret;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin >> s[0] >> s[1];
  for (int id = 0; id < 2; id++) {
    a[id][0] = (s[id][0] == 'A');
    for (int i = 1; i < (int)s[id].size(); i++) {
      a[id][i] = a[id][i - 1];
      if (s[id][i] == 'A')
        a[id][i]++;
      else
        a[id][i] = 0;
    }
    b[id][0] = (s[id][0] != 'A');
    for (int i = 1; i < (int)s[id].size(); i++) {
      b[id][i] = b[id][i - 1];
      b[id][i] += (s[id][i] != 'A');
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a--;
    b--;
    c--;
    d--;
    pair<int, int> p0 = query(0, a, b);
    pair<int, int> p1 = query(1, c, d);
    int ans = 1;
    if (p0.first > p1.first)
      ans = 0;
    else if (p0.first % 2 != p1.first % 2)
      ans = 0;
    if (!p0.first and p1.first) p0.second--;
    if (p0.second < p1.second) ans = 0;
    if (p0.first == p1.first and p0.second % 3 != p1.second % 3) ans = 0;
    cout << ans;
  }
  cout << endl;
  return 0;
}
