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;
long long a[2001], r[2001], F[30][2001];
void find_coeff(int t, int k) {
  memset(F[t], 0, sizeof(F[t]));
  if (k == 0) return;
  if (k == 1)
    for (int i = 0; i < N; i++) F[t][i] = 1;
  else {
    find_coeff(t + 1, k / 2);
    for (int i = 0; i < N; i++)
      for (int j = 0; j < i + 1; j++)
        F[t][i] = (F[t][i] + F[t + 1][j] * F[t + 1][i - j]) % 1000000007ll;
    if (k % 2 == 1) {
      memset(r, 0, sizeof(r));
      for (int i = 0; i < N; i++)
        for (int j = 0; j < i + 1; j++) r[i] = (r[i] + F[t][j]) % 1000000007ll;
      for (int i = 0; i < N; i++) F[t][i] = r[i];
    }
  }
}
int main() {
  cin >> N >> K;
  for (int i = 0; i < N; i++) cin >> a[i];
  find_coeff(0, K);
  memset(r, 0, sizeof(r));
  if (K == 0)
    for (int i = 0; i < N; i++) r[i] = a[i];
  else
    for (int i = 0; i < N; i++)
      for (int j = 0; j < i + 1; j++)
        r[i] = (r[i] + (long long)(a[j]) * F[0][i - j]) % 1000000007ll;
  for (int i = 0; i < N; i++) cout << r[i] << (i == N - 1 ? '\n' : ' ');
  return 0;
}
