A thief made his way to a shop.

As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is ai.

The thief is greedy, so he will take exactly k products (it's possible for some kinds to take several products of that kind).

Find all the possible total costs of products the thief can nick into his knapsack.

Input

The first line contains two integers n and k (1 ≤ n, k ≤ 1000) — the number of kinds of products and the number of products the thief will take.

The second line contains n integers ai (1 ≤ ai ≤ 1000) — the costs of products for kinds from 1 to n.

Output

Print the only line with all the possible total costs of stolen products, separated by a space. The numbers should be printed in the ascending order.

Examples

Input

3 2
1 2 3


Output

2 3 4 5 6


Input

5 5
1 1 1 1 1


Output

5


Input

3 3
3 5 11


Output

9 11 13 15 17 19 21 25 27 33
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = ((res % p) * (x % p)) % p;
    y = y >> 1;
    x = ((x % p) * (x % p)) % p;
  }
  return res;
}
long long raichu(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = ((res) * (x));
    y = y >> 1;
    x = ((x) * (x));
  }
  return res;
}
bool isprime(long long n) {
  if (n < 2)
    return false;
  else if (n == 2)
    return true;
  else if (n % 2 == 0)
    return false;
  else {
    long long z = sqrt(n);
    for (long long i = 0; i < z - 1; i++)
      if (n % (i + 2) == 0) return false;
    return true;
  }
}
const long long prime = 13313;
namespace FFT {
const long long N = 20;
const long long MAXN = (1 << N);
const double PI = acos(-1.0);
class cmplx {
 private:
  double x, y;

 public:
  cmplx() : x(0.0), y(0.0) {}
  cmplx(double a) : x(a), y(0.0) {}
  cmplx(double a, double b) : x(a), y(b) {}
  double get_real() { return this->x; }
  double get_img() { return this->y; }
  cmplx conj() { return cmplx(this->x, -(this->y)); }
  cmplx operator=(const cmplx& a) {
    this->x = a.x;
    this->y = a.y;
    return *this;
  }
  cmplx operator+(const cmplx& b) {
    return cmplx(this->x + b.x, this->y + b.y);
  }
  cmplx operator-(const cmplx& b) {
    return cmplx(this->x - b.x, this->y - b.y);
  }
  cmplx operator*(const double& num) {
    return cmplx(this->x * num, this->y * num);
  }
  cmplx operator/(const double& num) {
    return cmplx(this->x / num, this->y / num);
  }
  cmplx operator*(const cmplx& b) {
    return cmplx(this->x * b.x - this->y * b.y, this->y * b.x + this->x * b.y);
  }
  cmplx operator/(const cmplx& b) {
    cmplx temp(b.x, -b.y);
    cmplx n = (*this) * temp;
    return n / (b.x * b.x + b.y * b.y);
  }
};
cmplx w[MAXN];
cmplx f1[MAXN];
long long rev[MAXN];
void ReserveBits(long long k) {
  static long long rk = -1, lim;
  if (k == rk) return;
  rk = k, lim = 1 << k;
  for (long long i = 1; i <= lim; ++i) {
    long long j = rev[i - 1], t = k - 1;
    while (t >= 0 && ((j >> t) & 1)) {
      j ^= 1 << t;
      --t;
    }
    if (t >= 0) {
      j ^= 1 << t;
      --t;
    }
    rev[i] = j;
  }
}
void fft(cmplx* a, long long k) {
  ReserveBits(k);
  long long n = 1 << k;
  for (long long i = 0; i < n; ++i)
    if (rev[i] > i) swap(a[i], a[rev[i]]);
  for (long long l = 2, m = 1; l <= n; l += l, m += m) {
    if (w[m].get_real() == 0 && w[m].get_img() == 0) {
      double angle = PI / m;
      cmplx ww(cos(angle), sin(angle));
      if (m > 1) {
        for (long long j = 0; j < m; ++j) {
          if (j & 1)
            w[m + j] = w[(m + j) / 2] * ww;
          else
            w[m + j] = w[(m + j) / 2];
        }
      } else
        w[m] = cmplx(1, 0);
    }
    for (long long i = 0; i < n; i += l) {
      for (long long j = 0; j < m; ++j) {
        cmplx v = a[i + j], u = a[i + j + m] * w[m + j];
        a[i + j] = v + u;
        a[i + j + m] = v - u;
      }
    }
  }
}
vector<long long> multiply(const vector<long long>& a,
                           const vector<long long>& b) {
  long long k = 1;
  while ((1 << k) < (a.size() + b.size())) ++k;
  long long n = (1 << k);
  for (long long i = 0; i < n; ++i) f1[i] = cmplx(0, 0);
  for (long long i = 0; i < a.size(); ++i) f1[i] = f1[i] + cmplx(a[i], 0);
  for (long long i = 0; i < b.size(); ++i) f1[i] = f1[i] + cmplx(0, b[i]);
  fft(f1, k);
  for (long long i = 0; i <= n / 2; ++i) {
    cmplx p = f1[i] + f1[(n - i) % n].conj();
    cmplx _q = f1[(n - i) % n] - f1[i].conj();
    cmplx q(_q.get_img(), _q.get_real());
    f1[i] = (p * q) * 0.25;
    if (i > 0) f1[(n - i)] = f1[i].conj();
  }
  for (long long i = 0; i < n; ++i) f1[i] = f1[i].conj();
  fft(f1, k);
  vector<long long> ans(a.size() + b.size() - 1);
  for (long long i = 0; i < ans.size(); ++i) {
    ans[i] = (long long)(f1[i].get_real() / n + 0.5);
  }
  return ans;
}
};  // namespace FFT
using namespace FFT;
vector<long long> pfft(vector<long long>& a, long long k) {
  vector<long long> res;
  res.push_back(1);
  while (k > 0) {
    if (k & 1) res = multiply(res, a);
    k = k >> 1;
    a = multiply(a, a);
    for (long long i = 0; i < (long long)a.size(); i++)
      if (a[i] > 0) a[i] = 1;
    for (long long i = 0; i < (long long)res.size(); i++)
      if (res[i] > 0) res[i] = 1;
  }
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<long long> z(1001, 0);
  for (long long i = 0; i < n; i++) z[a[i]] = 1;
  vector<long long> ans = pfft(z, k);
  for (long long i = 0; i < (long long)ans.size(); i++)
    if (ans[i] > 0) cout << i << " ";
  return 0;
}
