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;
vector<long long> mul(vector<long long> a, vector<long long> b) {
  int n = a.size();
  vector<long long> ans(n);
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      ans[i] += (a[j] * b[n - 1 - j + i]) % 1000000007;
      ans[i] %= 1000000007;
    }
  }
  return ans;
}
vector<long long> pot(vector<long long> a, long long b) {
  int n = a.size();
  if (b == 0) {
    vector<long long> ans(n);
    ans[n - 1] = 1;
    return ans;
  }
  vector<long long> aux = pot(a, b / 2);
  aux = mul(aux, aux);
  if (b % 2 == 1) aux = mul(aux, a);
  return aux;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<long long> a(n, 1);
  vector<long long> b = pot(a, k);
  vector<long long> c(n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i + 1; j++) {
      c[i] += (v[j] * b[n - 1 - i + j]) % 1000000007;
      c[i] %= 1000000007;
    }
  cout << c[0];
  for (int i = 1; i < n; i++) cout << " " << c[i];
  cout << endl;
}
