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;
long long A[2010];
vector<vector<long long> > mult(vector<vector<long long> > A,
                                vector<vector<long long> > B) {
  long long n = A.size();
  vector<vector<long long> > C(n, vector<long long>(n, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      C[0][i] = (C[0][i] + (B[j][i] * A[0][j]) % 1000000007) % 1000000007;
    }
  }
  for (int i = 1; i < n; i++)
    for (int j = i; j < n; j++) C[i][j] = C[i - 1][j - 1];
  return C;
}
vector<vector<long long> > expo(vector<vector<long long> > A, long long k) {
  long long n = A.size();
  vector<vector<long long> > B(n, vector<long long>(n, 0));
  for (long long i = 0; i < n; i++) B[i][i] = 1;
  while (k) {
    if (k & 1) B = mult(B, A);
    A = mult(A, A);
    k /= 2;
  }
  return B;
}
long long rs(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> A[i];
  vector<vector<long long> > trans(n, vector<long long>(n, 0));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) trans[j][i] = 1;
  }
  vector<vector<long long> > res = expo(trans, k);
  for (int i = 0; i < n; i++) {
    long long ans = 0;
    for (int j = 0; j < n; j++)
      ans = (ans + (A[j] * res[j][i]) % 1000000007) % 1000000007;
  }
  cout << '\n';
}
