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 n, k, a[2001], s[2001], F[2001][10001];
const int base = 1e9 + 7;
long long IndianMul(long long a, long long b) {
  if (b == 0) return 0;
  long long t = IndianMul(a, b / 2);
  if ((b & 1) == 0) t = (t + t);
  if ((b & 1) == 1) t = (t + t + a);
  return t;
}
long long C(long long n, long long k) {
  if (F[n][k] != -1) return F[n][k];
  if (k == 0 || n == k) return F[n][k] = 1;
  return F[n][k] = C(n - 1, k - 1) + C(n - 1, k);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= k; j++) {
      F[i][j] = -1;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 1; j--) {
      s[i] += IndianMul(C(k + i - j - 1, k - 1), a[j]);
    }
    cout << s[i] << " ";
  }
}
