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;
int dx4[4] = {0, -1, 0, 1};
int dy4[4] = {1, 0, -1, 0};
int dx8[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int dy8[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
int Power(int base, int exp) {
  int ans = 1;
  while (exp > 0) {
    if (exp & 1) ans = (1LL * ans * base) % 1000000007;
    exp = exp >> 1;
    base = (1LL * base * base) % 1000000007;
  }
  return ans;
}
void input() {
  long long int n, k;
  cin >> n >> k;
  long long int a[n];
  for (int i = 0; i < n; i += 1) cin >> a[i];
  if (k == 0) {
    for (int i = 0; i < n; i += 1) cout << a[i] << ' ';
    cout << '\n';
    return;
  }
  for (int i = 0; i < n; i += 1) {
    long long int c = 1, ans = 0;
    for (int j = 0; j < i + 1; j += 1) {
      ans = (ans + (c * a[i - j]) % 1000000007) % 1000000007;
      c = (c * ((k + j) % 1000000007)) % 1000000007;
      c = (c * Power(j + 1, 1000000007 - 2)) % 1000000007;
    }
    cout << ans << ' ';
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  while (q--) input();
}
