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 a[10000];
long long mod = 1000000007;
long long ans[10000];
long long c[10000];
long long k, n;
long long ext_gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long resu = ext_gcd(b, a % b, x, y);
  long long temp = x;
  x = y;
  y = temp - a / b * y;
  return resu;
}
long long Inv(long long a, long long b) {
  long long resu, temp;
  ext_gcd(a, b, resu, temp);
  resu = (resu % b + b) % b;
  return resu;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  c[0] = 1;
  for (long long i = 1; i < n; i++)
    c[i] = c[i - 1] * (i + k - 1) % mod * Inv(i, mod) % mod;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j <= i; j++) {
      ans[i] += a[i - j] * c[j] % mod;
    }
  for (int i = 0; i < n; i++) cout << ans[i] % mod << " ";
  cout << endl;
  return 0;
}
