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 long long mod = 1000000007LL;
int n, k;
long long A[2020], v[2020], T[2020];
void mult(long long A[2020], long long v[2020], long long r[2020]) {
  long long aux[2020];
  for (int i = 0; i < n; i++) {
    aux[i] = 0LL;
    for (int j = 0; j <= i; j++) {
      if (!j)
        aux[i] = (aux[i] + A[i] * v[j]) % mod;
      else
        aux[i] = (aux[i] + A[i - j] * v[j]) % mod;
    }
  }
  for (int i = 0; i < n; i++) r[i] = aux[i];
  return;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> A[i];
  T[0] = 1LL;
  for (int i = 1; i < n; i++) T[i] = 0LL;
  for (int i = 0; i < n; i++) v[i] = 1LL;
  for (; k; k /= 2) {
    if (k & 1) mult(T, v, T);
    mult(v, v, v);
  }
  mult(T, A, A);
  for (int i = 0; i < n; i++) {
    if (i) cout << " ";
    cout << A[i];
  }
  cout << endl;
  return 0;
}
