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 ipower(long long int num, long long int n) {
  if (n == 0) return 1;
  long long int t = ipower(num, n / 2);
  if (n % 2 == 1)
    return (((t * t) % 1000000007) * num) % 1000000007;
  else
    return (t * t) % 1000000007;
}
int main() {
  int i, j, k, n;
  while (cin >> n) {
    cin >> k;
    long long int ans[n], a[n], coef[n];
    for (i = 0; i < n; i++) cin >> a[i];
    coef[0] = 1;
    for (i = 1; i < n; i++) {
      coef[i] = coef[i - 1];
      coef[i] *= (k + i - 1);
      coef[i] %= 1000000007;
      coef[i] *= ipower(i, 1000000007 - 2);
      coef[i] %= 1000000007;
    }
    for (i = 0; i < n; i++) {
      ans[i] = 0;
      for (j = i, k = 0; j >= 0; j--, k++) {
        ans[i] += (coef[k] * a[j]);
        ans[i] %= 1000000007;
      }
    }
    for (i = 0; i < n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}
