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;
long long MOD = 1e9 + 7;
long long MAX = 9223372036854775807;
string a, b;
vector<int> s, t;
bool f(int x, int y, int x2, int y2) {
  auto sl = lower_bound(s.begin(), s.end(), x);
  auto sr = upper_bound(s.begin(), s.end(), y);
  sr--;
  int sb = 0, sa = 0, tb = 0, ta = 0;
  int temp;
  if (sl == s.end() or *sl > y) {
    sa = y - x + 1;
  } else {
    sb = 1 + (int)(sr - sl);
    sa = y - (*sr);
  }
  auto tl = lower_bound(t.begin(), t.end(), x2);
  auto tr = upper_bound(t.begin(), t.end(), y2);
  tr--;
  if (tl == t.end() or *tl > y2) {
    ta = y2 - x2 + 1;
  } else {
    tb = 1 + (int)(tr - tl);
    ta = y2 - (*tr);
  }
  if (tb < sb or (tb - sb) % 2 != 0) return 0;
  if (ta > sa) return 0;
  temp = (sa - ta) % 3;
  int k = (tb - sb) / 2;
  if (k == 0) {
    if (temp == 0) return 1;
    return 0;
  }
  if (sb == 0) {
    if (sa == ta) return 0;
    return 1;
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  ;
  cin >> a >> b;
  for (int i = 0; i < a.size(); ++i) {
    if (a[i] == 'B')
      a[i] = 'C';
    else if (a[i] == 'C')
      a[i] = 'B';
    if (a[i] != 'A') s.push_back(i);
  }
  for (int i = 0; i < b.size(); ++i) {
    if (b[i] == 'B')
      b[i] = 'C';
    else if (b[i] == 'C')
      b[i] = 'B';
    if (b[i] != 'A') t.push_back(i);
  }
  int q;
  cin >> q;
  while (q--) {
    int r[4];
    for (int i = 0; i < 4; ++i) {
      cin >> r[i];
      r[i]--;
    }
    cout << f(r[0], r[1], r[2], r[3]);
    ;
  }
  cout << '\n';
  return 0;
}
