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 n;
int q[2005];
long long f[2005], m;
long long Pow(long long a, int k) {
  long long c = 1;
  while (k) {
    if (k & 1) c = c * a % 1000000007;
    k >>= 1;
    a = a * a % 1000000007;
  }
  return c;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> q[i];
  m--;
  f[0] = 1;
  for (int i = 1; i < n; i++)
    f[i] =
        f[i - 1] * (m + i) % 1000000007 * Pow(i, 1000000007 - 2) % 1000000007;
  for (int i = 0; i < n; i++) {
    long long s = 0;
    for (int j = 0; j <= i; j++) {
      s += q[j] * (f[i - j]) % 1000000007;
      if (s > 1000000007) s %= 1000000007;
    }
    if (i) cout << " ";
    cout << s;
  }
  cout << endl;
}
