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 mod = 1000 * 1000 * 1000 + 7;
long long pw(long long a, long long i) {
  if (i == 0ll) return 1;
  return (pw((a % mod) * (a % mod) % mod, i / 2) * (i & 1 ? a : 1)) % mod;
}
int const MAXN = 2 * 1000 + 5;
long long sa[MAXN][MAXN];
long long a[MAXN];
long long n, k;
void s(int i) {
  if (i == n + 1) return;
  for (int j = 1; j < i; j++) {
    sa[i][j] = (sa[i - 1][j] * ((k + (i - j - 1)) % mod)) % mod;
    sa[i][j] *= pw((i - j), mod - 2);
    sa[i][j] %= mod;
  }
  sa[i][i] = 1;
  s(i + 1);
  return;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sa[1][1] = 1;
  s(2);
  long long ans;
  for (int i = 1; i <= n; i++) {
    ans = 0;
    for (int j = 1; j <= n; j++) {
      ans += (sa[i][j] * a[j]) % mod;
      ans %= mod;
    }
    cout << ans << " ";
  }
}
