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 N = 2e3 + 77, Mod = 1e9 + 7;
int a[N], n, k, fact[N], rfact[N], p[N];
map<pair<int, int>, int> dp;
int Pow(int x, int y) {
  if (y == 0) return 1;
  int c = Pow(x, y / 2);
  c = (c * 1ll * c) % Mod;
  return (y & 1) ? (c * 1ll * x) % Mod : c;
}
int C(int x, int y) {
  if (x < 0) return 0;
  if (dp[make_pair(x, y)]) return dp[make_pair(x, y)];
  int c = 1;
  for (int i = y; i > x; i--) {
    c = (c * 1ll * i) % Mod;
  }
  c = (c * 1ll * rfact[y - x]) % Mod;
  dp[make_pair(x, y)] = c;
  dp[make_pair(y - x, y)] = c;
  return c;
}
int main() {
  ios_base::sync_with_stdio(false);
  fact[0] = rfact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = (fact[i - 1] * 1ll * i) % Mod;
    rfact[i] = Pow(fact[i], Mod - 2);
  }
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    dp[make_pair(k - 1, i + k - 1)] = C(k - 1, i + k - 1);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      p[j] =
          (p[j] * 1ll + a[i] * 1ll * dp[make_pair(k - 1, j - i + k - 1)]) % Mod;
    }
    a[i] = (a[i] + p[i]) % Mod;
    cout << a[i] << ' ';
  }
  return 0;
}
