Problem

There is a bundle of $ n $ cards, with $ i $ written on the $ i $ th card from the bottom. Shuffle this bundle as defined below $ q $ times.


* When looking at the bundle from below, consider a bundle $ x $ made by stacking odd-numbered cards on top, and a bundle $ y $ made by stacking even-numbered cards on top. Then do whatever you like out of the following $ 2 $.
* Operation $ 0 $: Put the bundle $ y $ on the bundle $ x $ to make a bundle of $ 1 $.
* Operation $ 1 $: Put the bundle $ x $ on the bundle $ y $ to make a bundle of $ 1 $.



For example, if you consider shuffling $ 1 $ times for a bundle of $ 6 $ cards (123456 from the bottom), the bundle you get when you perform the operation $ 0 $ is ~~ 246135 ~~ from the bottom. 135246, the bundle obtained by performing the operation $ 1 $ is ~~ 135246 ~~ 246135 from the bottom. (Corrected at 14:52)

After $ q $ shuffle, I want the card with $ k $ to be $ d $ th from the bottom of the bunch.
If possible, print $ q $ of shuffle operations in sequence, and if not, print $ -1 $.

Constraints

The input satisfies the following conditions.

* $ 2 \ le n \ le 10 ^ {15} $
* $ 1 \ le q \ le 10 ^ 6 $
* $ 1 \ le k, d \ le n $
* $ n $ is even

Input

The input is given in the following format.


$ n $ $ q $ $ k $ $ d $


Output

If it is possible to shuffle the conditions, output the shuffle operation on the $ q $ line.

If not possible, output $ -1 $.

Examples

Input

4 2 1 1


Output

0
0


Input

4 2 3 1


Output

0
1


Input

4 1 1 4


Output

-1


Input

7834164883628 15 2189823423122 5771212644938


Output

0
1
1
1
1
1
0
1
0
1
0
0
0
0
0
#include <bits/stdc++.h>
#define REP(i, a, n) for (ll i = ((ll) a); i < ((ll) n); i++)
using namespace std;
typedef long long ll;

bool check(vector<pair<ll, ll>> &v, ll x) {
  REP(i, 0, v.size()) {
    if (v[i].first <= x && x < v[i].second) {
      return true;
    }
  }
  return false;
}

int main(void) {
  ll N, Q, K, D;
  cin >> N >> Q >> K >> D, K--, D--;

  vector<vector<pair<ll, ll>>> z(Q + 1);
  z[Q].push_back(make_pair(D, D + 1));

  for (ll i = Q - 1; i >= 0; i--) {
    vector<pair<ll, ll>> v;
    REP(k, 0, z[i + 1].size()) {
      ll l = z[i + 1][k].first, h = z[i + 1][k].second;
      if (l < N / 2 && h <= N / 2) {
        v.push_back(make_pair(l * 2, h * 2));
      } else if (l < N / 2 && h > N / 2) {
        v.push_back(make_pair(l * 2, N));
        v.push_back(make_pair(0, h * 2 - N));
      } else if (l >= N / 2 && h > N / 2) {
        v.push_back(make_pair(l * 2 - N, h * 2 - N));
      }
    }

    sort(v.begin(), v.end());

    REP(k, 0, v.size()) {
      assert(v[k].first < v[k].second);
      if (z[i].size() >= 1) {
        if (z[i].back().second >= v[k].first) {
          z[i].back().second = v[k].second;
          continue;
        }
      }
      z[i].push_back(v[k]);
    }
  }

  if (!check(z[0], K)) {
    cout << -1 << endl;
  } else {
    vector<ll> ans(Q);
    ll x = K;
    REP(i, 0, Q) {
      ll n = x % 2 == 0 ? x / 2 : x / 2 + N / 2;
      if (check(z[i + 1], n)) {
        x = n;
        ans[i] = 0;
      } else {
        x = x % 2 == 0 ? x / 2 + N / 2 : x / 2;
        ans[i] = 1;
      }
    }
    REP(i, 0, Q) printf("%ld\n", ans[i]);
  }
}
