Problem: While most students still sit their exams, the tractor college has completed the summer exam session. In fact, students study only one subject at this college — the Art of Operating a Tractor. Therefore, at the end of a term a student gets only one mark, a three (satisfactory), a four (good) or a five (excellent). Those who score lower marks are unfortunately expelled.

The college has n students, and oddly enough, each of them can be on scholarship. The size of the scholarships varies each term. Since the end-of-the-term exam has just ended, it's time to determine the size of the scholarship to the end of next term.

The monthly budget for the scholarships of the Tractor college is s rubles. To distribute the budget optimally, you must follow these rules:

  * The students who received the same mark for the exam, should receive the same scholarship;
  * Let us denote the size of the scholarship (in roubles) for students who have received marks 3, 4 and 5 for the exam, as k3, k4 and k5, respectively. The values k3, k4 and k5 must be integers and satisfy the inequalities 0 ≤ k3 ≤ k4 ≤ k5;
  * Let's assume that c3, c4, c5 show how many students received marks 3, 4 and 5 for the exam, respectively. The budget of the scholarship should be fully spent on them, that is, c3·k3 + c4·k4 + c5·k5 = s;
  * Let's introduce function <image> — the value that shows how well the scholarships are distributed between students. In the optimal distribution function f(k3, k4, k5) takes the minimum possible value. 



Given the results of the exam, and the budget size s, you have to find the optimal distribution of the scholarship.

Input

The first line has two integers n, s (3 ≤ n ≤ 300, 1 ≤ s ≤ 3·105) — the number of students and the budget size for the scholarship, respectively. The second line contains n integers, where the i-th number represents the mark that the i-th student got for the exam. It is guaranteed that at each mark was given to at least one student.

Output

On a single line print three integers k3, k4 and k5 — the sought values that represent the optimal distribution of the scholarships. If there are multiple optimal answers, print any of them. If there is no answer, print -1.

Examples

Input

5 11
3 4 3 5 5


Output

1 3 3


Input

6 15
5 3 3 4 4 5


Output

-1

Solution (in CPP):

Solution:
#include<bits/stdc++.h> /*
#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp> // Including */
using namespace std;
//using namespace __gnu_pbds;

//typedefs

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int,int> pii;
typedef pair<double, double> pdd;
typedef pair<ll, ll> pll;
typedef vector<pii> vii;
typedef vector<pll> vll;
typedef vector<int>::iterator vit;
typedef set<int>::iterator sit;

/*
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename F, typename S>
using ordered_map = tree<F, S, less<F>, rb_tree_tag, tree_order_statistics_node_update>;
*/

//#Defines
#define rep(i,a,b)	for(i=a;i<=b;i++)
#define repR(i,a,b)	for(i=a;i>=b;i--)
//#define pb push_back
#define pb emplace_back
#define F first
#define S second
#define mp make_pair
#define all(c)		c.begin(),c.end()
#define endl '\n'
#define pf printf
#define sf scanf
//#define left __left
//#define right __right
//#define tree __tree
#define MOD 1000000007
//#define harmonic(n) 0.57721566490153286l+log(n)

#define RESET(a,b)	memset(a,b,sizeof(a))
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define sqr(a) ((a) * (a))

#define optimize() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define fraction() cout.unsetf(ios::floatfield); cout.precision(10); cout.setf(ios::fixed,ios::floatfield);

const double PI = acos(-1);
const double eps = 1e-9;
const int inf = 2000000000;
const ll infLL = 9000000000000000000;


//Bit Operations
inline bool checkBit(ll n, int i) { return n&(1LL<<i); }
inline ll setBit(ll n, int i) { return n|(1LL<<i);; }
inline ll resetBit(ll n, int i) { return n&(~(1LL<<i)); }

int fx[] = {0, 0, +1, -1};
int fy[] = {+1, -1, 0, 0};
//int dx[] = {+1, 0, -1, 0, +1, +1, -1, -1};
//int dy[] = {0, +1, 0, -1, +1, -1, +1, -1};


//Inline functions

inline bool EQ(double a, double b) { return fabs(a-b) < 1e-9; }
inline bool isLeapYear(ll year) { return (year%400==0) || (year%4==0 && year%100!=0); }
inline void normal(ll &a) { a %= MOD; (a < 0) && (a += MOD); }
inline ll modMul(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a*b)%MOD; }
inline ll modAdd(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); return (a+b)%MOD; }
inline ll modSub(ll a, ll b) { a %= MOD, b %= MOD; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while(p) { if(p&1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, MOD-2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }
inline bool isInside(pii p,ll n,ll m)	{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<m); }
inline bool isInside(pii p,ll n)		{ return (p.first>=0&&p.first<n&&p.second>=0&&p.second<n); }
inline bool isSquare(ll x) 				{ ll s = sqrt(x);	return (s*s==x); }
inline bool isFib(ll x) 				{ return isSquare(5*x*x+4)|| isSquare(5*x*x-4); }
inline bool isPowerOfTwo(ll x)			{ return ((1LL<<(ll)log2(x))==x); }


struct func
{
	//this is a sample overloading function for sorting stl
	bool operator()(pii const &a, pii const &b)	
	{
		if(a.F==b.F)
			return (a.S<b.S);
		return (a.F<b.F);
	}
};

//Prime Number Generator

/*
#define M 100000000
int marked[M/64 + 2];

#define on(x) (marked[x/64] & (1<<((x%64)/2)))
#define mark(x)  marked[x/64] |= (1<<((x%64)/2))

vl prime;

bool isPrime(int num) {
  return num > 1 && (num == 2 || ((num & 1) && !on(num))); 
}

void sieve(ll n) {
	for (ll i = 3; i * i < n; i += 2) {
		if (!on(i)) {
			for (ll j = i * i; j <= n; j += i + i) {
				mark(j);
			}
		}
	}
	prime.pb(2);
	for(ll i = 3; i <= n; i += 2) {
		if(!on(i))	prime.pb(i);	
	}
}

*/

//
//debug
#ifdef tajir
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p ) {
	return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v ) {
	os << "{";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
			os << *it;
	}
	return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
			os << *it;
	}
	return os << "]";
}

template < typename T >
ostream &operator << ( ostream & os, const multiset< T > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
		os << *it;
	}
	return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
	os << "[";
	for(auto it = v.begin(); it != v.end(); ++it) {
		if( it != v.begin() ) os << ", ";
		os << it -> first << " = " << it -> second ;
		}
	return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu () {
	cerr << endl;
}

template <typename T>
void faltu( T a[], int n ) {
	for(int i = 0; i < n; ++i) cerr << a[i] << ' ';
	cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) {
	cerr << arg << ' ';
	faltu(rest...);
}
#else
#define dbg(args...)
#endif // tajir

int c[6];
int n,s;

bool isOk(int x , int p , int q) {
	return ((q- c[4] * x) / c[5] )  >= x;
}

int f(int k[] ) {

	return abs(c[3] * k[3] - c[4] * k[4]) + abs(c[4] * k[4] - c[5] * k[5] );
}

bool checkCost(int k[]) {
	return c[3] * k[3] + c[4] * k[4] + c[5] * k[5] == s;
}

int main()
{
	#ifdef tajir
		freopen("input.txt", "r", stdin);
	#else
		// online submission
	#endif
	optimize();
	
	cin >> n >> s;

	for(int i = 0; i < n; ++i) {
		int x;
		cin >> x;
		c[x]++;
	}
//	dbg(c,6);
	int k1,k2,k3, pre = inf;

	int ans = inf;

	k1 = -1 , k2 = -1 , k3 = -1;

	int k[6];

	for(k[3] = (s/n) ; k[3] >= 0; --k[3]) {
		for(k[4] = ( s - c[3] * k[3]) / ( c[4] + c[5] ) ; k[4] >= k[3]; --k[4]) {
			k[5] = ( s - ( c[3] * k[3] + c[4] * k[4] ) ) / c[5];
			if ( c[5]*k[5]-c[4]*k[4] >= ans )
				break;
			dbg(checkCost(k));

			if(checkCost(k) ) {
				int now = abs(c[3] * k[3] - c[4] * k[4]) + abs(c[4] * k[4] - c[5] * k[5]);
				if(now < ans) {
					k1 = k[3];
					k2 = k[4];
					k3 = k[5];
					ans = now;
				}
			}
		}
	}

	if(ans == inf) {
		cout << -1 << endl;
		return 0;
	}

	cout << k1 << " " << k2 << " " << k3 << endl;
	
	return 0;
}






//?