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;
auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
long long modinv(long long x) {
  long long r = 1, y = 1000000007 - 2;
  for (; y > 0; y >>= 1, x = x * x % 1000000007)
    if (y & 1) r = r * x % 1000000007;
  return r;
}
int32_t main() {
  long long n;
  cin >> n;
  long long k;
  cin >> k;
  vector<long long> vec(n);
  for (long long i = 0; i < n; i++) cin >> vec[i];
  vector<long long> dp(n);
  dp[0] = 1;
  long long fact = 1;
  for (long long d = 1; d < n; d++) {
    fact = fact * d % 1000000007;
    dp[d] = 1;
    for (long long i = 0; i < d; i++)
      dp[d] = dp[d] * (d + k - 1 - i) % 1000000007;
    dp[d] = dp[d] * modinv(fact) % 1000000007;
  }
  vector<long long> ans(n);
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++) {
      long long d = j - i;
      (NULL);
      ans[j] += dp[d] * vec[i];
      ans[j] %= 1000000007;
    }
  }
  for (long long i = 0; i < n; i++) {
    if (i) cout << " ";
    cout << ans[i];
  }
  cout << endl;
}
