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 INF = 1000000111;
const int LIM = 10000010;
const int S_N = 100001;
const int MOD = 1000000007;
int n, k;
void mult(vector<int> a, vector<int> b, vector<int>& c) {
  vector<int> _(n);
  for (int i = 0; i < (n); ++i) {
    long long sum = 0;
    for (int j = 0; j <= (i); ++j) {
      sum += a[j] * (long long)b[i - j];
      if (sum >= 1000000 * MOD) sum %= MOD;
    }
    _[i] = sum;
  }
  c = _;
}
ostream& operator<<(ostream& out, vector<int> v) {
  for (int i = 0; i < (v.size()); ++i) out << v[i] << " ";
  out << endl;
  return out;
}
int main() {
  cin >> n >> k;
  vector<int> s(n);
  for (int i = 0; i < (n); ++i) cin >> s[i];
  vector<int> a(n, 0);
  a[0] = 1;
  vector<int> b(n, 1);
  for (k; k > 0; k /= 2) {
    if (k % 2) mult(a, b, a);
    mult(b, b, b);
  }
  mult(a, s, s);
  cout << s;
  return 0;
}
