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;
const int MAXN = 2001;
long long N, K;
long long B[MAXN];
long long S[MAXN];
long long A[MAXN];
long long pot(long long b, int e) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    long long r = pot(b, e / 2);
    return (r * r) % MOD;
  }
  return (pot(b, e - 1) * b) % MOD;
}
long long invMod(long long x) { return pot(x, MOD - 2); }
void FB() {
  for (int k = 1; k <= K; k++)
    for (int i = 2; i <= N; i++) A[i] = (A[i - 1] + A[i]) % MOD;
  for (int i = 1; i <= N; i++) cout << A[i] << ' ';
}
int main() {
  cin >> N >> K;
  for (int i = 1; i <= N; i++) cin >> A[i];
  if (K == 0) {
    for (int i = 1; i < N; i++) cout << A[i] << ' ';
    cout << A[N];
    return 0;
  }
  B[1] = 1;
  for (int n = 2; n <= N; n++) {
    B[n] = ((B[n - 1] * (K + n - 2)) % MOD) * (invMod(n - 1)) % MOD;
  }
  for (int i = N; i > 0; i--)
    for (int j = 1; j <= i; j++)
      S[i] = (S[i] + A[j] * B[N - j + 1 - (N - i)]) % MOD;
  for (int i = 1; i < N; i++) cout << S[i] << ' ';
  cout << S[N] << '\n';
  return 0;
}
