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 mod = 1000000000 + 7;
const int maxn = 2000 + 100;
long long ny[maxn], mul[maxn], f[maxn];
int n, k, a[maxn];
long long g(long long p, int n) {
  if (n == 0) return 1;
  if (n == 1) return p % mod;
  long long tmp = g(p, n / 2);
  tmp = (tmp * tmp) % mod;
  if (n & 1) tmp = (tmp * p) % mod;
  return tmp;
}
int main() {
  long long tmp;
  ny[1] = 1;
  for (int i = 2; i != maxn; i++) ny[i] = g(i, mod - 2);
  while (cin >> n >> k) {
    f[0] = 1;
    for (int i = 1; i <= n; i++) {
      f[i] = (((f[i - 1] * (k + i - 1)) % mod) * ny[i]) % mod;
    }
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) {
      tmp = 0;
      for (int j = 1; j <= i; j++) {
        tmp = (tmp + ((f[i - j] * a[j]) % mod)) % mod;
      }
      cout << tmp << " ";
    }
    cout << endl;
  }
  return 0;
}
