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 "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"
#include "numeric"
#include "cassert"

using namespace std;

const long long int MOD = 1000000007;
//const int MOD = 998244353;

long long int N, M, K, H, W, L, R;
//int N, M, K, H, W, L, R

long long int power(long long int x, long long int n, long long int M, long long int d) {
	long long int tmp = d;

	while (n) {
		if (n & 1) {
			tmp *= x;
			tmp %= M;
		}
		x *= x;
		x %= M;
		n /= 2;
	}
	return tmp;
}


int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	long long int n, q, k, d;
	cin >> n >> q >> k >> d;
	k--;
	d--;
	long long int LMOD = d;
	for (int i = 0; i < q; i++) {
		LMOD *= 2;
		LMOD %= n;
	}
	if (q < 55) {
		if (n > 1LL << q) {
			//long long int LMOD = d;
			long long int RMOD = LMOD + (1LL << q) - 1;
			//cout << LMOD << " " << k << " " << RMOD << endl;
			if (!((LMOD <= k && k <= RMOD) || (LMOD <= k + n && k + n <= RMOD))) {
				cout << -1 << endl;
				return 0;
			}

		}
	}
	//cout << bitset<15>(k - LMOD) << endl;
	long long int box = (k - LMOD + n) % n;
	vector<long long int>ans, tapi({ d });
	for (int i = 0; i < q; i++) {
		//if (d >= N / 2) {
		//}
		//else {

		//}
		d *= 2;
		d += q - 1 - i > 63 ? 0 : ((box >> q - 1 - i) & 1);
		d %= n;
		tapi.push_back(d);
		///cout << d << endl;
		//box /= 2;
	}
//	cout << d << " " << k << endl;
	assert(d == k);
	reverse(tapi.begin(), tapi.end());
	for (int i = 0; i < q; ++i) {
		ans.push_back((tapi[i] & 1) ^ (tapi[i + 1] >= n / 2));
	}
	for (auto i : ans)cout << i << endl;
}
