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 = 1e9 + 7;
const int MAXN = 2e3 + 15;
int a[MAXN], ans[MAXN], ch[MAXN];
inline int bPow(int x, int k) {
  int ret = 1;
  while (k) {
    if (k & 1) ret = 1LL * ret * x % MOD;
    x = 1LL * x * x % MOD;
    k >>= 1;
  }
  return ret;
}
inline int chose(int n, int k) {
  int ret = 1;
  for (int i = 1; i <= n; i++)
    ret = 1LL * ret * (k + i) % MOD * bPow(i, MOD - 2) % MOD;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) ch[i] = chose(i - 1, k - 1);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
      ans[j] = (ans[j] + 1LL * ch[j - i + 1] * a[i] % MOD) % MOD;
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
