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;
long long k;
long long arr[2010];
long long res[2010];
long long modP(long long x, long long y, long long z) {
  long long ret = 1;
  while (y) {
    if (y & 1) ret = (ret * x) % z;
    x = (x * x) % z;
    y >>= 1;
  }
  return ret;
}
int C(long long a, long long b) {
  if (a == 0 || a == b) return 1;
  if (a > b) return 0;
  long long ret = 1;
  for (int i = b; i > b - a; i--) ret = (ret * i) % 1000000007LL;
  for (int i = a; i > 0; i--)
    ret = (ret * modP(i, 1000000007LL - 2, 1000000007LL)) % 1000000007LL;
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> arr[i];
  for (int i = 0; i < n; i++)
    for (int j = i; j < n; j++)
      res[j] = (res[j] + arr[j - i] * C(i, i + k - 1)) % 1000000007LL;
  for (int i = 0; i < n; i++) cout << res[i] << ' ';
  cout << endl;
  return 0;
}
