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;
const int MAXN = 1e5 + 1;
const long long mod = 1e9 + 7;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long modInverse(long long n) { return power(n, mod - 2); }
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<long long> ans(n, 0);
  for (int i = 0; i < n; i++) {
    long long cur = 1;
    ans[i] = (ans[i] + (arr[i] * cur) % mod) % mod;
    for (int j = 1; j <= i; j++) {
      cur = (cur * (k + j - 1)) % mod;
      cur = (cur * modInverse(j)) % mod;
      ans[i] = (ans[i] + (arr[i - j] * cur) % mod) % mod;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  for (int I = 1; I <= t; I++) {
    solve();
  }
  return 0;
}
