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 N = 2e3 + 10, NX = 1e9 + 7;
int n, k, r, A[N], C[N], P[N], I[N];
int Power(int a, int b) {
  if (b == 0) return (1);
  int tmp = Power(a, b / 2);
  tmp = (1ll * tmp * tmp) % NX;
  if (b & 1) tmp = (1ll * tmp * a) % NX;
  return (tmp);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &A[i]);
  I[0] = C[0] = r = 1;
  for (int i = 1; i <= n; i++) I[i] = (1ll * I[i - 1] * i) % NX;
  I[n] = Power(I[n], NX - 2);
  for (int i = n - 1; i >= 1; i--) I[i] = (1ll * I[i + 1] * (i + 1)) % NX;
  for (int i = 1; i < n; i++)
    r = (1ll * r * (k + i - 1)) % NX, C[i] = (1ll * r * I[i]) % NX;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++) P[i] = (P[i] + 1ll * C[i - j] * A[j]) % NX;
  for (int i = 0; i < n; i++) printf("%d ", P[i]);
  return (0);
}
