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;
long long me(long long x, long long y) {
  if (x == 1) return 1;
  if (y == 0) return 1;
  if (y % 2)
    return (x * me(x, y - 1)) % (long long)(1e9 + 7);
  else
    return me((x * x) % (long long)(1e9 + 7), y / 2) % (long long)(1e9 + 7);
}
long long inv(long long x) { return me(x, (long long)(1e9 + 7) - 2); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n, k;
  cin >> n >> k;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long values[n + 1];
  if (k == 0) {
    for (int i = 0; i < n; i++) cout << a[i] << " ";
    return 0;
  }
  values[0] = 1;
  long long cur_k = k - 1, cur_r = 0;
  for (int i = 1; i <= n; i++) {
    cur_k++;
    cur_r++;
    values[i] =
        ((((cur_k)*inv(cur_r)) % (long long)(1e9 + 7)) * values[i - 1]) %
        (long long)(1e9 + 7);
  }
  vector<int> ans(n);
  for (int i = 0; i < n; i++) {
    long long value = 0;
    for (int j = i; j >= 0; j--) {
      value = (value + (a[j] * values[i - j]) % (long long)(1e9 + 7)) %
              (long long)(1e9 + 7);
    }
    ans[i] = value;
  }
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
}
