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 = 1e5 + 5;
string s, t;
int ns, nt;
int ps[maxn], pt[maxn];
int goods[maxn], goodt[maxn];
int ans[maxn];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> s;
  cin >> t;
  ns = s.length();
  nt = t.length();
  for (int i = (0); i < (ns); i++) {
    ps[i] = (i == 0) ? 0 : ps[i - 1];
    if (s[i] != 'A') ++ps[i];
  }
  for (int i = (0); i < (nt); i++) {
    pt[i] = (i == 0) ? 0 : pt[i - 1];
    if (t[i] != 'A') ++pt[i];
  }
  int la = -1;
  for (int i = (0); i < (ns); i++) {
    if (s[i] != 'A') la = i;
    goods[i] = la;
  }
  la = -1;
  for (int i = (0); i < (nt); i++) {
    if (t[i] != 'A') la = i;
    goodt[i] = la;
  }
  int q;
  cin >> q;
  for (int i = (0); i < (q); i++) {
    int ls, rs, lt, rt;
    cin >> ls >> rs >> lt >> rt;
    --ls;
    --rs;
    --lt;
    --rt;
    int cnts = ps[rs] - (ls == 0 ? 0 : ps[ls - 1]);
    int cntt = pt[rt] - (lt == 0 ? 0 : pt[lt - 1]);
    if (cnts > cntt) {
      ans[i] = 0;
      continue;
    }
    if (cnts % 2 != cntt % 2) {
      ans[i] = 0;
      continue;
    }
    int da = (cntt - cnts) / 2;
    int ss, st;
    ss = rs + 1 - max(ls, goods[rs]);
    st = rt + 1 - max(lt, goodt[rt]);
    bool can_make = false;
    if (ss >= st && (da >= 1 || (da == 0 && ss % 3 == st % 3))) can_make = true;
    if (can_make)
      ans[i] = 1;
    else
      ans[i] = 0;
  }
  for (int i = (0); i < (q); i++) cout << ans[i];
  return 0;
}
