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;
int n, K, size;
int v[2010];
long long C[2010];
long long sol[2010];
long long ModN(long long x) {
  long long a = x % 1000000007;
  if (a < 0) return (a + 1000000007);
  return a;
}
void Euclid_extins(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  } else {
    Euclid_extins(b, a % b, x, y);
    int aux, q = a / b;
    aux = x;
    x = y;
    y = aux - y * q;
  }
}
inline void Combinari() {
  long long i, j, f, inv, aux;
  for (i = 1; i <= n; i++) {
    C[i] = 1LL;
    for (j = K; j <= i + K - 2; j++) {
      C[i] *= j;
      C[i] %= 1000000007;
    }
    f = 1LL;
    for (j = 2; j < i; j++) {
      f *= j;
      f %= 1000000007;
    }
    inv = aux = 0LL;
    Euclid_extins(f, 1000000007, inv, aux);
    inv = ModN(inv);
    C[i] *= inv;
    C[i] %= 1000000007;
  }
}
int main() {
  int i, j;
  cin >> n >> K;
  for (i = 1; i <= n; i++) cin >> v[i];
  if (K == 0) {
    for (i = 1; i <= n; i++) cout << v[i] << ' ';
    cout << "\n";
    return 0;
  }
  Combinari();
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= i; j++) {
      sol[i] += C[i - j + 1] * (long long)v[j];
      sol[i] %= 1000000007;
    }
  }
  for (i = 1; i <= n; i++) cout << sol[i] << ' ';
  cout << "\n";
  return 0;
}
