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 a[2001], b[2001], f[2001];
int n, k;
const int mod = 1e9 + 7;
int pw(int x, int y, int z) {
  if (!y) return 1;
  int w = pw(x, y / 2, z);
  w = 1ll * w * w % z;
  if (y & 1) return 1ll * w * x % z;
  return w;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (!k) {
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    return 0;
  }
  k--;
  f[0] = 1;
  for (int i = 1; i <= n; i++) {
    f[i] = 1ll * (1ll * f[i - 1] * (k + i)) * pw(i, mod - 2, mod) % mod;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      b[j] = (b[j] + 1ll * f[j - 1] * a[i]) % mod;
    }
  }
  for (int i = 1; i <= n; i++) printf("%d ", b[i]);
  return 0;
}
