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;
long long n, k, i, j, b[1010], a[1010], ans[1010];
long long p(long long b, long long e) {
  long long ret = 1;
  while (e) {
    if (e & 1) ret = (ret * b) % 1000000007;
    b = (b * b) % 1000000007;
    e >>= 1;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  b[1] = 1;
  k--;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 2; i <= n; i++)
    b[i] = (b[i - 1] * (k + i - 1) * p(i - 1, 1000000007 - 2)) % 1000000007;
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++) ans[j] += a[i] * b[j - i + 1];
  for (i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}
