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;
long long int n, k, a[(2005)], ans[(2005)], t;
long long int fe(long long int x, long long int k) {
  if (!k) return 1;
  long long int t = fe(x, k / 2);
  return k & 1 ? t * t % ((int)1e9 + 7) * x % ((int)1e9 + 7)
               : t * t % ((int)1e9 + 7);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  t = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++)
      ans[j] = (ans[j] + a[j - i + 1] * t) % ((int)1e9 + 7);
    t = t * (k + i - 1) % ((int)1e9 + 7) * fe(i, ((int)1e9 + 7) - 2) %
        ((int)1e9 + 7);
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}
