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 div(int x) {
  long long res = 1;
  long long tmp = x;
  int y = 1000000007 - 2;
  while (y > 0) {
    if (y % 2 == 1) {
      res *= tmp;
      res %= 1000000007;
    }
    y /= 2;
    tmp *= tmp;
    tmp %= 1000000007;
  }
  return res;
}
long long Com(int n, int k) {
  long long res = 1;
  for (int i = 1; i <= k; ++i) {
    res = res * (n - i + 1);
    res %= 1000000007;
    res = res * div(i);
    res %= 1000000007;
  }
  return res;
}
int main() {
  long long n, k;
  long long a[2005];
  long long s[2005];
  long long C[2005];
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    s[i] = 0;
  }
  for (int i = 0; i <= n; ++i) {
    C[i] = Com(k - 1 + i, i);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      s[i] += a[j] * C[i - j];
      s[i] %= 1000000007;
    }
  }
  for (int i = 0; i < n - 1; ++i) cout << s[i] << " ";
  cout << s[n - 1] << "\n";
  return 0;
}
