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;
int n, k;
int c[2222];
class Matrix {
 public:
  int mat[2222];
  void init() { memset(mat, 0, sizeof(mat)); }
  void setOne() {
    for (int i = 1; i <= n; i++) mat[i] = 1;
  }
  Matrix mul(Matrix a, Matrix b) {
    Matrix c;
    c.init();
    for (int i = 1; i <= n; i++) {
      int k = i - 1;
      for (int j = 1; j <= n - i + 1; j++) {
        k++;
        c.mat[k] = (c.mat[k] + ((long long)a.mat[i] * b.mat[j] % 1000000007)) %
                   1000000007;
      }
    }
    return c;
  }
  Matrix pw(Matrix a, int k) {
    if (k == 1) return a;
    Matrix c = pw(a, k / 2);
    c = mul(c, c);
    if (k & 1) c = mul(c, a);
    return c;
  }
  void out() {
    for (int i = 1; i <= n; i++) cout << mat[i] << " ";
    cout << "\n";
  }
};
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> c[i];
  Matrix a;
  a.setOne();
  if (k == 0) {
    for (int i = 1; i <= n; i++) cout << c[i] << " ";
    return 0;
  }
  a = a.pw(a, k);
  for (int kk = 1; kk <= n; kk++) {
    int ret = 0;
    for (int j = 1, i = kk; j <= kk; j++, i--)
      ret = (ret + ((long long)c[j] * a.mat[i]) % 1000000007) % 1000000007;
    cout << ret << " ";
  }
  return 0;
}
