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;
class Solution {
 public:
  vector<int> answer(const string& s, const string& t,
                     vector<vector<int>>& qs) {
    int q = qs.size();
    int n = s.length();
    int m = t.length();
    vector<int> res;
    res.reserve(q);
    vector<int> cnts1(n + 1, 0);
    vector<int> pre1(n + 1, -1);
    vector<int> cnts2(m + 1, 0);
    vector<int> pre2(m + 1, -1);
    for (int i = 0, j = -1; i < n; ++i) {
      cnts1[i + 1] = cnts1[i];
      if (s[i] == 'B' || s[i] == 'C') {
        ++cnts1[i + 1];
        j = i;
      }
      pre1[i] = j;
    }
    for (int i = 0, j = -1; i < m; ++i) {
      cnts2[i + 1] = cnts2[i];
      if (t[i] == 'B' || t[i] == 'C') {
        ++cnts2[i + 1];
        j = i;
      }
      pre2[i] = j;
    }
    for (auto& qr : qs) {
      int al = qr[0], ar = qr[1];
      int bl = qr[2], br = qr[3];
      int c1 = cnts1[ar + 1] - cnts1[al];
      int c2 = cnts2[br + 1] - cnts2[bl];
      int a1 = ar - max(al - 1, pre1[ar]);
      int a2 = br - max(bl - 1, pre2[br]);
      int ans = 0;
      if (c1 <= c2 && (((c1 ^ c2) & 0x1) == 0)) {
        if (a1 >= a2) {
          if (c1 == c2 && (a1 - a2) % 3 == 0) {
            ans = 1;
          } else if (c1 < c2) {
            if (c1 > 0 || (c1 == 0 && a1 > a2)) {
              ans = 1;
            }
          }
        }
      }
      res.push_back(ans);
    }
    return res;
  }
};
int main(int argc, char** argv) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  string s, t;
  cin >> s;
  cin >> t;
  int n;
  cin >> n;
  vector<vector<int>> qs(n, vector<int>(4, 0));
  for (int i = 0; i < n; ++i) {
    cin >> qs[i][0] >> qs[i][1] >> qs[i][2] >> qs[i][3];
    --qs[i][0];
    --qs[i][1];
    --qs[i][2];
    --qs[i][3];
  }
  Solution sol;
  auto res = sol.answer(s, t, qs);
  for (auto r : res) {
    cout << r;
  }
  cout << endl;
}
