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;
int INF = 1000000007;
int gcd(int a, int b, int &x, int &y) {
  if (!a) {
    x = 0;
    y = 1;
    return b;
  }
  int x1, y1;
  int d = gcd(b % a, a, x1, y1);
  x = y1 - x1 * (b / a);
  y = x1;
  return d;
}
int res(int b, int a) {
  int x, y, d;
  d = gcd(a, INF, x, y);
  if (x < 0) x += INF;
  return (1ll * x * b) % INF;
}
int main() {
  int i, j, n, k, help;
  cin >> n >> k;
  vector<int> v(n), arr(n);
  for (i = 0; i < n; i++) cin >> v[i];
  if (!k) {
    for (i = 0; i < n; i++) cout << v[i] << " ";
    return 0;
  }
  arr[0] = 1;
  for (i = 1; i < n; i++) {
    help = (1ll * arr[i - 1] * (k - 1 + i)) % INF;
    arr[i] = res(help, i);
  }
  for (i = 0; i < n; i++) {
    help = 0;
    for (j = i + 1; j > 0; j--)
      help = (1ll * help + 1ll * arr[i - j + 1] * v[j - 1]) % INF;
    cout << help << " ";
  }
  return 0;
}
