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 r = 1000 * 1000 * 1000 + 7;
const int G = 2000 + 10;
long long a[G], ans[G], cmb[G];
long long ent(int n, int k) {
  long long ans = 1;
  for (int i = n; i > k; i--) {
    ans *= i;
    ans = ans % r;
  }
  return ans % r;
}
long long pw(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  long long c = pw(a, b / 2);
  c = (1LL * c * c) % r;
  if (b % 2) c = (c * a) % r;
  return c;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  if (k == 0)
    for (int i = 1; i <= n; i++) ans[i] = a[i];
  else {
    for (int j = 1; j <= n; j++) {
      long long e = ent(k + j - 2, k - 1);
      long long t = ent(j - 1, 1);
      long long invn = pw(t, r - 2);
      cmb[j] = (invn * e) % r;
    }
    for (int i = 1; i <= n; i++) {
      int y = i;
      for (int j = 1; j <= i; j++) {
        ans[i] += (a[y] * (cmb[j])) % r;
        y--;
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] % r << ' ';
  return 0;
}
