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 maxn = 2010, mod = 1e9 + 7, mod2 = 1e7, x = 103;
long long t[maxn], arr[maxn];
int powmod(int a, int b) {
  if (b == 0) return 1;
  int x = powmod(a, b / 2);
  x = x * (long long)x % mod;
  if (b & 1) x = x * (long long)a % mod;
  return x;
}
int f(int a, int b) {
  b--;
  b += a;
  a = b - a;
  int ret = 1;
  for (int i = a + 1; i <= b; i++)
    ret = ret * (long long)i % mod * powmod(i - a, mod - 2) % mod;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (register int i = (0); i < (int)(n); ++i) cin >> arr[i];
  for (register int i = (0); i < (int)(n); ++i) t[i] = f(i, k);
  if (k == 0)
    for (register int i = (0); i < (int)(n); ++i) cout << arr[i] << ' ';
  else
    for (register int i = (0); i < (int)(n); ++i) {
      long long ans = 0;
      for (register int j = (0); j < (int)(i + 1); ++j)
        ans = (ans + (t[j] * arr[i - j])) % mod;
      cout << ans << ' ';
    }
  cout << endl;
  return 0;
}
