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;
const long long mod = 1e9 + 7;
long long n, k, a[2020], inv[2020], c[2020];
signed main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; ++i) cin >> a[i];
  if (!k) {
    for (long long i = 1; i <= n; ++i) cout << a[i] << ' ';
    return 0;
  }
  inv[0] = inv[1] = 1;
  for (long long i = 2; i <= n; ++i)
    inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  c[0] = 1;
  for (long long i = 1; i <= n; ++i)
    c[i] = c[i - 1] * inv[i] % mod * (i + k - 1) % mod;
  for (long long i = 1, ans = 0; i <= n; ++i, ans = 0) {
    for (long long j = 1; j <= i; ++j) (ans += a[j] * c[i - j] % mod) %= mod;
    cout << ans << ' ';
  }
  return 0;
}
