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;
const int maxn = 2e3 + 100;
const int mod = 1e9 + 7;
long long int d[maxn], a[maxn], c[maxn];
long long int p(int x, int y) {
  if (x == 0) return 0;
  if (y == 0) return 1;
  long long int d = p(x, y / 2);
  if (y % 2 == 0) return ((d * d) % mod);
  d = d * d;
  d = d % mod;
  return ((d * x) % mod);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  c[0] = 1;
  for (int i = 1; i <= n; i++)
    c[i] = (c[i - 1] * (k + i - 1)) % mod * p(i, mod - 2) % mod;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++) d[i] = (d[i] + a[j] * c[i - j]) % mod;
  for (int i = 1; i <= n; i++) cout << d[i] << " ";
  cout << endl;
}
