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 a[2005];
long long inv[2005];
long long c[2005];
long long calC(int n, int m) {
  long long ans = 1;
  for (int i = 1; i <= m; ++i) {
    ans = ((ans * inv[i]) % 1000000007 * (n - i + 1)) % 1000000007;
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (k == 0) {
    for (int i = 0; i < n; i++) {
      printf("%lld ", a[i]);
    }
    printf("\n");
    return 0;
  }
  inv[0] = inv[1] = 1;
  for (int i = 2; i < 2005; i++) {
    inv[i] =
        ((-1000000007 / i * inv[1000000007 % i]) % 1000000007 + 1000000007) %
        1000000007;
  }
  for (int i = 0; i < n; i++) {
    c[i] = calC(i + k - 1, i);
  }
  for (int i = 0; i < n; i++) {
    long long ans = 0;
    for (int j = 0; j <= i; j++) {
      ans = (ans + a[j] * c[i - j]) % 1000000007;
    }
    cout << ans << " ";
  }
  printf("\n");
  return 0;
}
