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 = 1000000007;
long long inv(long long x) {
  long long y = mod - 2, z = 1;
  while (y) {
    if (y & 1) z = z * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return z;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k, a[2019] = {};
  long long b[2019] = {}, z[2019] = {};
  for (int i = 0; i <= 2000; i++) b[i] = inv(i);
  cin >> n >> k;
  if (k == 0) {
    for (int i = 0; i < n; i++) cin >> a[i], cout << a[i] << ' ';
    return 0;
  }
  k--;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    long long y = 1;
    for (int j = k + 1; j <= k + i; j++) y = y * j % mod;
    for (int j = 1; j <= i; j++) y = y * b[j] % mod;
    z[i] = y;
  }
  for (int i = 0; i < n; i++) {
    long long y = 0;
    for (int j = 0; j <= i; j++) {
      y = (y + z[i - j] * a[j]) % mod;
    }
    cout << y << ' ';
  }
}
