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>
#pragma comment(linker, "/stack:20000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")
using namespace std;
const int N = 1100;
const int MAXN = int(1e6) + 100;
const int Mmask = (1 << 20);
const int mod = int(1e9) + 7;
const long long MOD = (long long)(1e18) + 7ll;
int solution();
int main(int argc, char* const argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  return solution();
}
string s, t;
int p[MAXN], p1[MAXN], su[MAXN], su1[MAXN];
int solution() {
  cin >> s >> t;
  s = '$' + s;
  t = '$' + t;
  for (int i = 1; i < s.size(); i++)
    p[i] = p[i - 1] + (s[i] == 'B' || s[i] == 'C');
  for (int i = 1; i < t.size(); i++)
    p1[i] = p1[i - 1] + (t[i] == 'B' || t[i] == 'C');
  for (int i = 1; i < s.size(); i++) su[i] = (s[i] == 'A' ? su[i - 1] + 1 : 0);
  for (int i = 1; i < t.size(); i++)
    su1[i] = (t[i] == 'A' ? su1[i - 1] + 1 : 0);
  int q;
  cin >> q;
  while (q--) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    int prefs = p[b] - p[a - 1], preft = p1[d] - p1[c - 1];
    int suffs = min(su[b], b - a + 1), sufft = min(su1[d], d - c + 1);
    if (!prefs && preft && suffs == sufft)
      cout << 0;
    else {
      if ((prefs % 2) == (preft % 2) && prefs <= preft && suffs >= sufft) {
        int d = (suffs - sufft) % 3;
        if (d == 0)
          cout << 1;
        else {
          if (prefs + 2 <= preft)
            cout << 1;
          else
            cout << 0;
        }
      } else
        cout << 0;
    }
  }
  return 0;
}
