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 n;
vector<long long> operator*(vector<long long>& a, vector<long long>& b) {
  vector<long long> res(n, 0);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j)
      res[i] = (res[i] + a[j] * b[i - j] % 1000000007) % 1000000007;
  }
  return res;
}
vector<long long> pow(vector<long long>& b, int e) {
  if (e == 0) {
    vector<long long> id(n, 0);
    id[0] = 1;
    return id;
  }
  vector<long long> w = pow(b, e / 2);
  vector<long long> a = w * w;
  if (e % 2 == 1) return a * b;
  return a;
}
int main() {
  int k;
  cin >> n >> k;
  vector<long long> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  vector<long long> b(n, 1);
  vector<long long> res = pow(b, k);
  b = res * v;
  for (int i = 0; i < n; ++i) {
    if (i) cout << " ";
    cout << b[i] % 1000000007;
  }
  cout << endl;
}
