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>
char bin[] = "111011100110101100101000000101";
using namespace std;
long long int inver(int a) {
  long long int ans = 1, j = a;
  for (int i = 29; i >= 0; i--) {
    if (bin[i] == '1') {
      ans = (ans * j) % 1000000007;
    }
    j = (j * j) % 1000000007;
  }
  ans = (ans + 1000000007) % 1000000007;
  return ans;
}
int main() {
  int n, k, a[2222], ans[2222];
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long int sum, inv[2222];
  for (int i = 0; i < n; i++) ans[i] = 0;
  for (int i = 0; i < n; i++) {
    sum = 1;
    for (int j = 1; j < (n - i); j++) {
      if (!i) {
        inv[j] = inver(j);
      }
      sum = ((((sum * (k + j - 1)) % 1000000007) * inv[j]) % 1000000007);
      ans[j + i] = ((ans[j + i] + ((sum * a[i]) % 1000000007)) % 1000000007);
    }
    ans[i] = (ans[i] + a[i]) % 1000000007;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << endl;
  return 0;
}
