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;
long long n, k, a[2001], dp[2001];
long long power(long long a, long long p) {
  if (p == 0) return 1;
  long long temp = power(a, p / 2), ans = temp * temp;
  temp %= 1000000007;
  if (p % 2 == 1) ans *= a;
  ans %= 1000000007;
  return ans;
}
long long C(long long n, long long k) {
  if (n == k) return 1;
  long long up = 1, down = 1, ans;
  (i, l, r) for (long long i = l; i < r; i++)(i, 0, k) {
    up *= (n - i);
    up %= 1000000007;
    down *= (i + 1);
    down %= 1000000007;
  }
  ans = up * power(down, 1000000007 - 2);
  ans %= 1000000007;
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  (i, l, r) for (long long i = l; i < r; i++)(i, 0, n) cin >> a[i];
  if (!k) {
    (i, l, r) for (long long i = l; i < r; i++)(i, 0, n) cout << a[i] << " ";
    return 0;
  }
  (i, l, r) for (long long i = l; i < r; i++)(i, 1, n) dp[i] = C(i + k - 1, i);
  (i, l, r) for (long long i = l; i < r; i++)(i, 0, n) {
    long long ans = a[i];
    (i, l, r) for (long long i = l; i < r; i++)(j, 0, i) {
      ans += dp[i - j] * a[j];
    }
    cout << ans << " ";
  }
}
