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 long long inf = 1e18;
const long long maxn = 1e5 + 10;
const long long mod = 1e9 + 7;
#pragma GCC Optimize("Ofast")
long long sms[maxn], smt[maxn];
long long lsts[maxn], lstt[maxn];
int main() {
  string s, t;
  cin >> s >> t;
  for (long long i = 1; i <= (s).size(); i++) {
    lsts[i] = lsts[i - 1];
    if (s[i - 1] == 'B' || s[i - 1] == 'C') lsts[i] = i;
    sms[i] = sms[i - 1] + (s[i - 1] == 'B' || s[i - 1] == 'C');
  }
  for (long long i = 1; i <= (t).size(); i++) {
    lstt[i] = lstt[i - 1];
    if (t[i - 1] == 'B' || t[i - 1] == 'C') lstt[i] = i;
    smt[i] = smt[i - 1] + (t[i - 1] == 'B' || t[i - 1] == 'C');
  }
  long long q;
  cin >> q;
  string ans;
  while (q--) {
    long long a, b, c, d;
    cin >> a >> b >> c >> d;
    long long ones = sms[b] - sms[a - 1], onet = smt[d] - smt[c - 1];
    long long zeros = b - max(lsts[b], a - 1), zerot = d - max(lstt[d], c - 1);
    if (ones > onet || (ones % 2) != (onet % 2) || zeros < zerot) {
      ans += "0";
      continue;
    }
    if (ones == onet && (zeros % 3) != (zerot % 3)) {
      ans += "0";
      continue;
    }
    if (zeros == zerot && zeros == b - a + 1 && onet > 0) {
      ans += "0";
      continue;
    }
    ans += "1";
  }
  cout << ans;
}
