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 mod = 1e9 + 7;
int n;
struct Matrix {
  vector<long long> v = vector<long long>(n);
  Matrix operator*(const Matrix& other) {
    Matrix product;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= i; j++) {
        product.v[i] = (product.v[i] + v[j] * other.v[i - j]) % mod;
      }
    }
    return product;
  }
};
Matrix expo_power(Matrix a, long long k, int n) {
  Matrix product;
  product.v = a.v;
  while (k > 0) {
    if (k % 2) {
      product = product * a;
    }
    a = a * a;
    k /= 2;
  }
  return product;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long k;
  cin >> n >> k;
  vector<long long> init(n);
  for (int i = 0; i < n; i++) {
    cin >> init[i];
  }
  if (k == 0) {
    for (int i = 0; i < n; i++) {
      cout << init[i] << " ";
    }
    return 0;
  }
  Matrix mult;
  for (int i = 0; i < n; i++) {
    mult.v[i] = 1;
  }
  Matrix ans = expo_power(mult, k - 1, n);
  vector<vector<long long> > use(n, vector<long long>(n));
  for (int j = 0; j < n; j++) {
    use[0][j] = ans.v[j];
  }
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      use[i][j] = use[i - 1][j - 1];
    }
  }
  vector<long long> out(n);
  for (int j = 0; j < n; j++) {
    for (int i = 0; i < n; i++) {
      out[j] = (out[j] + (init[i] * use[i][j]) % mod) % mod;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << out[i] << " ";
  }
  cout << endl;
}
