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 mod = 1e9 + 7;
int gcd(int a, int b, int& x, int& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  int x1, y1;
  int d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
int inv(int a) {
  int x, y;
  int t = gcd(a, mod, x, y);
  return ((x % mod) + mod) % mod;
}
int main() {
  int n, k;
  cin >> n >> k;
  --k;
  vector<unsigned long long> co(n + 1);
  co[0] = 1;
  for (int i = 1; i <= n; ++i) {
    co[i] = (((co[i - 1] * (k + i) % mod) % mod) * inv(i)) % mod;
  }
  vector<long long> v(n + 1);
  for (int i = 1; i <= n; ++i) cin >> v[i];
  vector<long long> t = v;
  for (int i = 1; i <= n; ++i) {
    long long z = 1;
    for (int j = i + 1; j <= n; ++j) {
      t[j] += (v[i] * co[j - i]) % mod;
      t[j] %= mod;
    }
  }
  for (int i = 1; i <= n; ++i) cout << t[i] << " ";
}
