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, arr[3000], d[3000];
long long int fe(long long int a, long long int b) {
  if (!b) return 1;
  long long int t = fe(a, b / 2);
  t = t * t % 1000000007LL;
  if (b & 1) t = t * a % 1000000007LL;
  return t;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  d[1] = 1;
  for (int i = 2; i <= n; i++)
    d[i] = d[i - 1] * (k + i - 2) % 1000000007LL * fe(i - 1, 1000000007LL - 2) %
           1000000007LL;
  long long int sum;
  for (int i = 1; i <= n; i++) {
    sum = 0;
    for (int j = 1; j <= i; j++)
      sum = (sum + arr[j] * d[i - j + 1]) % 1000000007LL;
    cout << sum << " ";
  }
  cout << endl;
}
