You've got an array a, consisting of n integers. The array elements are indexed from 1 to n. Let's determine a two step operation like that:

  1. First we build by the array a an array s of partial sums, consisting of n elements. Element number i (1 ≤ i ≤ n) of array s equals <image>. The operation x mod y means that we take the remainder of the division of number x by number y. 
  2. Then we write the contents of the array s to the array a. Element number i (1 ≤ i ≤ n) of the array s becomes the i-th element of the array a (ai = si). 



You task is to find array a after exactly k described operations are applied.

Input

The first line contains two space-separated integers n and k (1 ≤ n ≤ 2000, 0 ≤ k ≤ 109). The next line contains n space-separated integers a1, a2, ..., an — elements of the array a (0 ≤ ai ≤ 109).

Output

Print n integers — elements of the array a after the operations are applied to it. Print the elements in the order of increasing of their indexes in the array a. Separate the printed numbers by spaces.

Examples

Input

3 1
1 2 3


Output

1 3 6


Input

5 0
3 14 15 92 6


Output

3 14 15 92 6
#include <bits/stdc++.h>
using namespace std;
long long n, k;
class matrix {
 public:
  long long d;
  long long a[2005];
  matrix(long long u) { d = u; }
  matrix operator*(matrix R) {
    matrix res = matrix(n);
    for (long long i = 0; i < n; i++) {
      res.a[i] = 0;
      for (long long j = 0; j <= i; j++) res.a[i] += a[j] * R.a[i - j];
    }
    return res;
  }
};
matrix init() {
  matrix res = matrix(n);
  for (long long i = 0; i < n; i++) res.a[i] = 1;
  return res;
}
matrix binpow(matrix mat, long long p) {
  if (p == 0) return init();
  if (p & 1) return mat * binpow(mat * mat, p / 2);
  return binpow(mat * mat, p / 2);
}
void solve() {
  cin >> n >> k;
  matrix arr = matrix(n);
  for (long long i = 0; i < n; i++) cin >> arr.a[i];
  matrix b = init();
  if (k) arr = binpow(b, k - 1) * arr;
  for (long long i = 0; i < n; i++) cout << arr.a[i] << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
