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;
long long gcd(long long a, long long b) { return b != 0 ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long extgcd(long long a, long long b, long long &x, long long &y) {
  long long g = a;
  x = 1;
  y = 0;
  if (b != 0) g = extgcd(b, a % b, y, x), y -= (a / b) * x;
  return g;
}
long long invMod(long long a, long long m) {
  long long x, y;
  if (extgcd(a, m, x, y) == 1)
    return (x + m) % m;
  else
    return 0;
}
const long double INF = 1e+100, EPS = 1e-14;
long long rem(long long a, long long b) { return (a % b + b) % b; }
long long quot(long long a, long long b) { return (a - rem(a, b)) / b; }
long long ceilI(long double);
long long floorI(long double x) {
  if (x > EPS)
    return (long long)(x + EPS);
  else if (x < -EPS)
    return -ceilI(-x);
  else
    return 0;
}
long long ceilI(long double x) {
  if (x > EPS)
    return (long long)(x + 1 - EPS);
  else if (x < -EPS)
    return -floorI(-x);
  return 0;
}
int n, s;
int c[10];
long long x, y, g, S;
long long c4, c5;
const long long INT_INF = 1000000000000000000ll;
long long mini = INT_INF;
long long a3, a4, a5;
void check2(long long k3, long long k4, long long k5) {
  assert(k3 * c[3] + k4 * c[4] + k5 * c[5] == s);
  if (!(k3 <= k4 and k4 <= k5)) return;
  long long val = abs(c[3] * k3 - c[4] * k4) + abs(c[4] * k4 - c[5] * k5);
  if (val < mini) mini = val, a3 = k3, a4 = k4, a5 = k5;
}
void check(long long k3, long double L, long double R) {
  long long X = x * S / g, Y = y * S / g;
  assert(c[4] * X + c[5] * Y == S);
  long double t2;
  long long t, k4, k5;
  t2 = (long double)(L - X) / (long double)c5;
  t = ceilI(t2);
  k4 = X + c5 * t, k5 = Y - c4 * t;
  assert(k4 * c[4] + k5 * c[5] == S);
  if (k4 < R + EPS and k5 >= 0) {
    check2(k3, k4, k5);
  }
  t2 = (long double)(R - X) / (long double)c5;
  t = floorI(t2);
  k4 = X + c5 * t, k5 = Y - c4 * t;
  assert(k4 * c[4] + k5 * c[5] == S);
  if (L - EPS < k4 and k5 >= 0) {
    check2(k3, k4, k5);
  }
}
int main() {
  memset(c, 0, sizeof(c));
  cin >> n >> s;
  for (int i = 0; i < (int)(n); ++i) {
    int a;
    cin >> a;
    c[a]++;
  }
  g = extgcd(c[4], c[5], x, y);
  c4 = c[4] / g;
  c5 = c[5] / g;
  for (int k3 = 0;; k3++) {
    S = (long long)s - k3 * c[3];
    if (S < 0) break;
    if (S % g != 0) continue;
    long double L = -INF, R = INF;
    L = max(L, (long double)k3);
    R = min(R, S / (long double)(c[4] + c[5]));
    do {
      long double L2 = L, R2 = R;
      R2 = min(R, (long double)c[3] * k3 / (long double)c[4]);
      L2 = max(L, (long double)S / (long double)(c[4] * 2));
      check(k3, L2, R2);
    } while (0);
    do {
      long double L2 = L, R2 = R;
      L2 = max(L, (long double)c[3] * k3 / (long double)c[4]);
      R2 = min(R, (long double)S / (long double)(c[4] * 2));
      check(k3, L2, R2);
    } while (0);
    do {
      long double L2 = L, R2 = R;
      R2 = min(R, (long double)c[3] * k3 / (long double)c[4]);
      R2 = min(R, (long double)S / (long double)(c[4] * 2));
      check(k3, L2, R2);
    } while (0);
    do {
      long double L2 = L, R2 = R;
      L2 = max(L, (long double)c[3] * k3 / (long double)c[4]);
      L2 = max(L, (long double)S / (long double)(c[4] * 2));
      check(k3, L2, R2);
    } while (0);
  }
  if (mini == INT_INF) {
    cout << -1 << endl;
  } else {
    cout << a3 << " " << a4 << " " << a5 << endl;
  }
  return 0;
}