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>
#pragma comment(linker, "/STACK:67108864")
using namespace std;
const int INF = 1000 * 1000 * 1000 + 5;
const int MOD = 1000 * 1000 * 1000 + 7;
const long double EPS = 1e-8;
int n, k;
vector<int> a;
vector<int> f, ans;
int pw(int a, int b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) res = (1ll * res * a) % MOD, --b;
    a = (1ll * a * a) % MOD, b /= 2;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  a.resize(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  f.resize(n);
  f[0] = 1;
  for (int i = 1; i < n; ++i)
    f[i] =
        (1ll * ((1ll * f[i - 1] * (k + i - 1)) % MOD) * pw(i, MOD - 2)) % MOD;
  ans.resize(n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= i; ++j)
      ans[i] = (ans[i] + ((1ll * f[i - j] * a[j]) % MOD)) % MOD;
  for (int i = 0; i < n; ++i) cout << ans[i] << " ";
  return 0;
}
