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 = 1e9 + 7;
int r[2000] = {0, 1};
int cnk(int n, int k) {
  if (k <= 0) return 1;
  int zn = 1;
  for (int i = 1; i <= k; ++i)
    zn = (((1LL * zn * (n - i + 1)) % mod) * r[i]) % mod;
  return zn;
}
int main() {
  int n, k, a[2000], c[2000];
  scanf("%d%d", &n, &k);
  for (int i = 2; i < n; ++i) r[i] = (mod - (mod / i) * r[mod % i] % mod) % mod;
  for (int i = 0; i < n; ++i) c[i] = cnk(k + n - i - 2, n - 1 - i);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    int res = 0;
    for (int j = 0; j <= i; ++j)
      res = (res + 1LL * c[n - 1 - i + j] * a[j]) % mod;
    printf("%d ", res);
  }
  printf("\n");
  return 0;
}
