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 N = 2009, mod = 1e9 + 7;
int n, k, C[N], a[N], inv[N], ans[N];
inline void Init() {
  inv[0] = inv[1] = 1;
  for (int i = 2; i < N; i++)
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
  C[0] = 1;
  for (int i = 1; i < N; i++)
    C[i] = 1ll * C[i - 1] * inv[i] % mod * (i + k - 1) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (!k) {
    {
      for (int i = 1; i <= n; i++) printf("%d ", a[i] % mod);
    }
    return 0;
  }
  Init();
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      ans[i] = (ans[i] + 1ll * a[j] * C[i - j] % mod) % mod;
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
  return 0;
}
