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 = 1e9 + 7;
const long long INFL = 1e18 + 123;
const double PI = atan2(0, -1);
mt19937 tw(960172);
long long rnd(long long x, long long y) {
  static uniform_int_distribution<long long> d;
  return d(tw) % (y - x + 1) + x;
}
struct value {
  int bc;
  int l;
  int na;
  value() {
    l = 0;
    bc = 0;
    na = 0;
  }
  value(char c) {
    l = 1;
    if (c == 'A') {
      bc = 0;
      na = 1;
    } else {
      bc = 1;
      na = 0;
    }
  }
  value operator+(value other) {
    value ret;
    ret.l = l + other.l;
    ret.bc = bc + other.bc;
    if (other.na == other.l) {
      ret.na = other.l + na;
    } else {
      ret.na = other.na;
    }
    return ret;
  }
};
int bpv;
vector<value> rsq;
value s_get(int v, int vl, int vr, int l, int r) {
  if (r <= vl || vr <= l) {
    return value();
  }
  if (l <= vl && vr <= r) {
    return rsq[v];
  }
  int vm = (vl + vr) / 2;
  return s_get(v * 2, vl, vm, l, r) + s_get(v * 2 + 1, vm, vr, l, r);
}
void solve() {
  string s;
  cin >> s;
  string t;
  cin >> t;
  string tot = s + t;
  bpv = 1;
  while (bpv < ((int)(tot).size())) {
    bpv *= 2;
  }
  rsq = vector<value>(bpv * 2);
  for (int i = 0; i < ((int)(tot).size()); ++i) {
    rsq[i + bpv] = value(tot[i]);
  }
  for (int i = bpv - 1; i; --i) {
    rsq[i] = rsq[i * 2] + rsq[i * 2 + 1];
  }
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int l1, r1, l2, r2;
    cin >> l1 >> r1 >> l2 >> r2;
    --l1;
    --l2;
    auto v1 = s_get(1, 0, bpv, l1, r1);
    auto v2 = s_get(1, 0, bpv, l2 + ((int)(s).size()), r2 + ((int)(s).size()));
    if (v1.bc % 2 != v2.bc % 2 || v1.bc > v2.bc) {
      cout << "0";
      continue;
    }
    if (v1.na < v2.na) {
      cout << "0";
      continue;
    }
    if (v1.bc == 0 && v2.bc && v1.na == v2.na) {
      cout << "0";
      continue;
    }
    int tmp = (v1.na - v2.na) % 3;
    if (tmp != 0 && v2.bc - v1.bc == 0) {
      cout << "0";
      continue;
    }
    cout << "1";
  }
  cout << "\n";
}
int main() {
  cerr << fixed << setprecision(15);
  cout << fixed << setprecision(15);
  ios::sync_with_stdio(false);
  int tests = 1;
  for (int it = 1; it <= tests; ++it) {
    solve();
  }
  return 0;
}
