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;
int GetNumber() {
  char c;
  while (c = getchar(), c < '0' || c > '9')
    ;
  int X = c - 48;
  while (c = getchar(), c >= '0' && c <= '9') X = X * 10 + c - 48;
  return X;
}
const int Mod = 1000000007;
long long Pow(long long A, int B) {
  if (!B) return 1;
  long long T = Pow(A * A % Mod, B / 2);
  if (B & 1) T = T * A % Mod;
  return T;
}
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  static long long A[2000];
  for (int i = 0; i < N; i++) A[i] = GetNumber();
  static long long B[2000];
  B[0] = 1;
  for (int i = 1; i < N; i++)
    B[i] = B[i - 1] * (K + i - 1) % Mod * Pow(i, Mod - 2) % Mod;
  static long long Ans[2000];
  memset(Ans, 0, sizeof(Ans));
  for (int i = 0; i < N; i++)
    for (int j = i; j < N; j++) Ans[j] = (Ans[j] + A[i] * B[j - i]) % Mod;
  for (int i = 0; i < N; i++) {
    printf("%d", (int)Ans[i]);
    putchar((i == N - 1) ? '\n' : ' ');
  }
  return 0;
}
