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>
using namespace std;
inline long long labs(long long x) { return (x > 0 ? x : -x); }
template <typename T>
inline T sqr(T x) {
  return x * x;
}
long long k3, k4, k5;
long long gcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
bool find_any_solution(long long a, long long b, long long c, long long& x0,
                       long long& y0, long long& g) {
  g = gcd(labs(a), labs(b), x0, y0);
  if (c % g != 0) return false;
  x0 *= c / g;
  y0 *= c / g;
  if (a < 0) x0 *= -1;
  if (b < 0) y0 *= -1;
  return true;
}
double func(long long c3, long long c4, long long c5) {
  long long res = labs(c3 * k3 - c4 * k4);
  res += labs(c4 * k4 - c5 * k5);
  return res;
}
bool found = false;
long long minval = INT_MAX;
long long cc3, cc4, cc5;
void dostuff(long long c3, long long c4, long long c5) {
  minval = func(c3, c4, c5);
  cc3 = c3;
  cc4 = c4;
  cc5 = c5;
  found = true;
}
int main() {
  long long n, s;
  cin >> n >> s;
  for (long long i = 0; i < (n); ++i) {
    char c;
    cin >> c;
    if (c == '3')
      k3++;
    else if (c == '4')
      k4++;
    else
      k5++;
  }
  long long k = k3 + k4 + k5;
  for (long long c3 = 0; c3 <= ceil((s + 0.0) / k); ++c3) {
    if (c3 == 3498) {
      ;
      ;
      long long a = 1;
      ;
      ;
      ;
    }
    long long s0 = s - c3 * k3;
    long long g = 0;
    long long c4, c5;
    if (!find_any_solution(k4, k5, s0, c4, c5, g)) continue;
    long long dc4 = k5 / g;
    long long dc5 = k4 / g;
    long long maxc5 = (s - c3 * k3 - c3 * k4) / k5;
    long long minc4 = c3;
    long long sdvig = ((c4 - c3) / dc4);
    if (c4 - sdvig * dc4 < c3) {
      sdvig++;
    } else {
      sdvig = min(sdvig, (maxc5 - c5) / dc5);
    }
    c5 += sdvig * dc5;
    c4 -= sdvig * dc4;
    if (c5 < c4) continue;
    long long K = (c5 - c4) / (dc4 + dc5) + 1;
    long long CC1 = k3 * c3 - c4 * k4;
    long long CC2 = c4 * k4 - k5 * c5;
    long long CC3 = dc4 * k4 + k5 * dc5;
    CC3 *= -1;
    long long CC4 = k4 * dc4;
    CC4 *= -1;
    long long g1 = CC1 / CC4;
    long long g2 = CC2 / CC3;
    vector<long long> vals;
    vals.push_back(0);
    vals.push_back(g1);
    vals.push_back(g1 + 1);
    vals.push_back(g2);
    vals.push_back(g2 + 1);
    vals.push_back(K - 1);
    for (long long i = 0; i < (vals.size()); ++i) {
      long long j = vals[i];
      if (j < K && j >= 0 && func(c3, c4 + j * dc4, c5 - j * dc5) < minval &&
          c3 <= c4 + j * dc4 && c4 + j * dc4 <= c5 - j * dc5)
        dostuff(c3, c4 + j * dc4, c5 - j * dc5);
    }
  }
  if (!found)
    cout << "-1" << endl;
  else
    cout << cc3 << " " << cc4 << " " << cc5 << endl;
  return 0;
}