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>
const long long mod = 1e9 + 7;
using namespace std;
const long long mod2 = 998244353;
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> coeff(n);
  coeff[0] = 1;
  long long prev_coeff = 1;
  for (int i = 1; i < n; i++) {
    long long temp = (prev_coeff * ((k + i - 1) % mod)) / (i);
    temp %= mod;
    coeff[i] = temp;
    prev_coeff = temp;
  }
  for (int i = 0; i < n; i++) {
    long long ans = 0;
    for (int j = 0; j <= i; j++) {
      ans += ((a[j] % mod) * (coeff[i - j]));
      ans %= mod;
    }
    cout << ans << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
