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 = 2010;
const long long mo = 1000000007;
typedef long long T1[2010][2010];
long long f[maxn], ans[maxn], g[maxn];
long long N, k;
long long x, y, q;
void extend_Eulid(long long a, long long b) {
  if (b == 0) {
    x = 1;
    y = 0;
    q = a;
  } else {
    extend_Eulid(b, a % b);
    int temp = x;
    x = y;
    y = temp - a / b * y;
  }
}
long long inv(long long a, long long n) {
  extend_Eulid(a, n);
  x = (x + mo) % mo;
  return x;
}
int main() {
  scanf("%I64d%I64d", &N, &k);
  for (int i = 1; i <= N; i++) scanf("%I64d", &g[i]);
  f[1] = 1;
  for (long long i = 2; i <= N; i++)
    f[i] = f[i - 1] * (k + i - 2) % mo * inv(i - 1, mo) % mo;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= i; j++)
      ans[i] += f[i - j + 1] * g[j], ans[i] = ans[i] % mo;
  if (k == 0)
    for (int i = 1; i <= N; i++) ans[i] = g[i];
  for (int i = 1; i <= N; i++) cout << ans[i] << ' ';
  cout << endl;
}
