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 maxn = 100500;
int sum1[maxn];
int sum2[maxn];
int cnt1[maxn];
int cnt2[maxn];
int main() {
  string s, t;
  cin >> s >> t;
  int n = s.size();
  int m = t.size();
  sum1[0] = 0;
  cnt1[0] = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i - 1] != 'A') {
      sum1[i] = sum1[i - 1] + 1;
      cnt1[i] = 0;
    } else {
      sum1[i] = sum1[i - 1];
      cnt1[i] = cnt1[i - 1] + 1;
    }
  }
  sum2[0] = 0;
  cnt2[0] = 0;
  for (int i = 1; i <= m; ++i) {
    if (t[i - 1] != 'A') {
      sum2[i] = sum2[i - 1] + 1;
      cnt2[i] = 0;
    } else {
      sum2[i] = sum2[i - 1];
      cnt2[i] = cnt2[i - 1] + 1;
    }
  }
  int q;
  cin >> q;
  for (int z = 0; z < q; ++z) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int a1, a2, x, y;
    x = sum1[b] - sum1[a - 1];
    y = sum2[d] - sum2[c - 1];
    a1 = min(cnt1[b], b - a + 1);
    a2 = min(cnt2[d], d - c + 1);
    if (x > y) {
      cout << 0;
      continue;
    }
    if (x == y) {
      if (a1 >= a2 && (a1 - a2) % 3 == 0) {
        cout << 1;
      } else {
        cout << 0;
      }
      continue;
    }
    if ((y - x) % 2 != 0) {
      cout << 0;
      continue;
    }
    if (a1 > a2) {
      cout << 1;
      continue;
    }
    if (a1 < a2) {
      cout << 0;
      continue;
    }
    if (x == 0) {
      cout << 0;
    } else {
      cout << 1;
    }
  }
  return 0;
}
