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, i, j, s, ar[2005], inv[2005], dp[2005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> ar[i];
  dp[0] = 1;
  dp[1] = k;
  inv[1] = 1;
  for (i = 2; i < n; i++) {
    inv[i] = ((long long int)(1e9 + 7) - (((long long int)(1e9 + 7) / i) *
                                          inv[(long long int)(1e9 + 7) % i]) %
                                             (long long int)(1e9 + 7)) %
             (long long int)(1e9 + 7);
    dp[i] = (((dp[i - 1] * (k + i - 1)) % (long long int)(1e9 + 7)) * inv[i]) %
            (long long int)(1e9 + 7);
  }
  for (i = 1; i <= n; i++) {
    s = 0;
    for (j = 0; j < i; j++)
      s = (s + (dp[j] * ar[i - j]) % (long long int)(1e9 + 7)) %
          (long long int)(1e9 + 7);
    cout << s << " ";
  }
  cout << "\n";
  return 0;
}
