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 MAX_N = 2e3 + 6, MOD = 1e9 + 7;
long long n, k;
long long Arr[MAX_N];
long long ans[MAX_N];
long long choose[MAX_N];
long long Pow(long long a, long long b) {
  if (!b) return 1;
  long long res = Pow(a, b >> 1);
  res = (res * res) % MOD;
  if (b & 1) res = (res * a) % MOD;
  return res;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie();
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> Arr[i];
  if (!k) {
    for (long long i = 0; i < n; i++) cout << Arr[i] << " ";
    return 0;
  }
  choose[0] = 1;
  for (int i = 1; i < MAX_N; i++)
    choose[i] = (choose[i - 1] * (i + k - 1)) % MOD * Pow(i, MOD - 2) % MOD;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j <= i; j++)
      ans[i] = (ans[i] + choose[i - j] * Arr[j]) % MOD;
  for (long long i = 0; i < n; i++) cout << ans[i] << " ";
  return 0;
}
