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 double PI = 3.14159265358979323846;
long long a[2000], r[2000], t[2000];
int n;
const int mod = 1000000007;
void mal(long long* x, long long* y) {
  memset(t, 0, sizeof(t));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      t[i] = (t[i] + x[j] * y[i - j]) % mod;
    }
  }
  for (int i = 0; i < n; i++) {
    x[i] = t[i];
  }
}
int main() {
  int k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) r[i]++;
  for (; k; k >>= 1) {
    if (k & 1) mal(a, r);
    mal(r, r);
  }
  cout << a[0];
  for (int i = 1; i < n; i++) {
    cout << ' ' << a[i];
  }
  cout << endl;
  return 0;
}
