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 long long mod = (1e9 + 7);
long long n, k, v[2050], dp[2050], ans[2050];
long long poww(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1; i <= n; i++) {
    dp[i] = 1LL;
    for (int j = i + 1; j <= n; j++) {
      long long delta = j - i - 1;
      dp[j] = ((dp[j - 1]) * (k + delta)) % mod;
      dp[j] = (dp[j] * poww(delta + 1, mod - 2)) % mod;
      ans[j] = (ans[j] + dp[j] * v[i]) % mod;
    }
    ans[i] = (ans[i] + v[i]) % mod;
  }
  for (int i = 1; i <= n; i++) cout << (ans[i]) % mod << " \n"[i == n];
}
