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 double pi = 3.1415926535897932384626433832795;
const int inf = 1e9;
const long long inf2 = 1e18;
const int mod = 1e9 + 7;
void boost() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
inline long long exp(long long x, long long n, long long m = mod) {
  long long r = 1;
  while (n) {
    if (n & 1) r = r * x % m;
    x = x * x % m, n >>= 1;
  }
  return r;
}
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> ncr(n + 1, 0);
  vector<int> arr(n + 1);
  for (int i = 1; i < n + 1; ++i) cin >> arr[i];
  ncr[0] = 1;
  for (int i = 1; i < n + 1; ++i) {
    ncr[i] = (ncr[i - 1] * 1LL * (k + i - 1)) % mod;
    ncr[i] = (ncr[i] * exp(i, mod - 2)) % mod;
  }
  vector<int> ans(n + 1, 0);
  for (int i = 1; i < n + 1; ++i) {
    for (int j = i; j < n + 1; ++j) {
      ans[j] = (ans[j] * 1LL + 1LL * arr[i] * ncr[j - i]) % mod;
    }
  }
  for (int i = 1; i < n + 1; ++i) cout << ans[i] << ' ';
  return;
}
int main() {
  boost();
  int tc = 1;
  while (tc--) solve();
  return 0;
}
