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;
const long long int M = 1e9 + 7;
long long int arr[2002], ans[2002];
long long int power(long long int a, long long int b) {
  long long int val = 1;
  while (b) {
    if (b % 2) val = (val * a) % M;
    a = (a * a) % M;
    b /= 2;
  }
  return val;
}
int main() {
  long long int n, k, i, j;
  long long int x, val = 0;
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (i = 1; i <= n; i++) {
    ans[i] = (ans[i] + arr[i]) % M;
    val = arr[i];
    x = 1;
    for (j = i + 1; j <= n; j++) {
      val = (val * (k + x - 1)) % M;
      val = (val * power(x, M - 2)) % M;
      x++;
      ans[j] = (ans[j] + val) % M;
    }
  }
  for (i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
