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 = (int)1e9;
const long long linf = (long long)1e18;
const int mod = (int)1e9 + 7;
const long double eps = (long double)1e-8;
const int maxn = (int)1e5 + 5;
const long double pi = acos(-1);
string s, t;
int as[maxn], bs[maxn], at[maxn], bt[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> s;
  cin >> t;
  memset((as), 0, sizeof(as));
  memset((bs), 0, sizeof(bs));
  memset((at), 0, sizeof(at));
  memset((bt), 0, sizeof(bt));
  for (auto i = 0; i < ((int)(s).size()); ++i) {
    as[i] += (s[i] == 'A');
    if (s[i] != 'A') as[i] = 0;
    bs[i] += (s[i] == 'B' || s[i] == 'C');
    as[i + 1] = as[i];
    bs[i + 1] = bs[i];
  }
  for (auto i = 0; i < ((int)(t).size()); ++i) {
    at[i] += (t[i] == 'A');
    if (t[i] != 'A') at[i] = 0;
    bt[i] += (t[i] == 'B' || t[i] == 'C');
    at[i + 1] = at[i];
    bt[i + 1] = bt[i];
  }
  string ans = "";
  int m;
  cin >> m;
  for (auto i = 0; i < m; ++i) {
    int ls, rs, lt, rt;
    cin >> ls >> rs >> lt >> rt;
    --ls;
    --rs;
    --lt;
    --rt;
    int kas = min(as[rs], rs - ls + 1);
    rs -= kas;
    int kat = min(at[rt], rt - lt + 1);
    rt -= kat;
    int ss = (rs < 0 ? 0 : bs[rs]) - (ls == 0 ? 0 : bs[ls - 1]);
    int st = (rt < 0 ? 0 : bt[rt]) - (lt == 0 ? 0 : bt[lt - 1]);
    int ost = (kas - kat) % 3;
    if (ss == 0) {
      if (st == 0) {
        if (kas >= kat && ost == 0)
          ans += '1';
        else
          ans += '0';
        continue;
      }
      ss += 2;
      kas -= 1;
      kas -= abs(kas - kat) % 3;
      ost = (kas - kat) % 3;
    }
    if (ss <= st && ss % 2 == st % 2 && kas >= kat &&
        (ost == 0 || (st - ss) >= 2))
      ans += '1';
    else
      ans += '0';
  }
  cout << ans << endl;
  return 0;
}
