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 int MOD = 1000000007;
const int MAX = 4025;
long long a[MAX], f[MAX], fa[MAX][MAX], rev[MAX], ans[MAX];
long long pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    b >>= 1;
    a *= a;
    ans %= MOD;
    a %= MOD;
  }
  return ans;
}
long long c(long long r) { return fa[r][r + 2005] * rev[r] % MOD; }
int main() {
  ios::sync_with_stdio(false);
  f[0] = 1;
  rev[0] = 1;
  for (int i = 1; i < MAX; i++) {
    f[i] = f[i - 1] * i % MOD;
    rev[i] = pow(f[i], MOD - 2);
  }
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < MAX; i++) fa[0][i] = 1;
  for (int i = 1; i < MAX; i++)
    for (int j = 1; j < MAX; j++)
      fa[i][j] = fa[i - 1][j - 1] * (j + k - 2006) % MOD;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = i; j <= n; j++) {
      ans[j] += a[i] * c(j - i);
      ans[j] %= MOD;
    }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
