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 N = 1e9 + 7;
long long a[2005];
int n, k;
vector<vector<long long> > mul(vector<vector<long long> > &A,
                               vector<vector<long long> > &B) {
  vector<vector<long long> > C(1, vector<long long>(n));
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++)
      C[0][i] = (C[0][i] + A[0][j] * B[0][i - j]) % N;
  return C;
}
vector<vector<long long> > pow(vector<vector<long long> > u, int k) {
  vector<vector<long long> > I(1, vector<long long>(n));
  I[0][0] = 1;
  while (k > 0) {
    if (k & 1) I = mul(I, u);
    u = mul(u, u);
    k >>= 1;
  }
  return I;
}
int main() {
  cin >> n >> k;
  int i, j;
  for (i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long> > A(1, vector<long long>(n));
  for (i = 0; i < n; i++) A[0][i] = 1;
  vector<vector<long long> > b(1, vector<long long>(n));
  b = pow(A, k);
  cout << a[0] % N << " ";
  long long temp = 0;
  for (i = 1; i < n; i++) {
    temp = 0;
    for (j = 0; j <= i; j++) temp = (temp + a[j] * b[0][i - j]) % N;
    cout << temp << " ";
  }
  cout << endl;
  return 0;
}
