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 MAX = 2005, MOD = 1e9 + 7;
long long A[MAX], T[MAX], C[MAX];
int n, m, l;
long long exp(long long a, long long b) {
  if (b == 0) return 1;
  if (b % 2) return ((a % MOD) * (exp(a, b - 1) % MOD)) % MOD;
  long long sq = exp(a, b / 2) % MOD;
  sq = (sq * sq) % MOD;
  return sq;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> A[i];
  if (m == 0) {
    for (int i = 1; i < n; i++) cout << A[i] << " ";
    cout << A[n] << "\n";
    return 0;
  }
  C[n] = 1;
  for (int i = n - 1; i > 0; i--) {
    int k = n - i + 1;
    C[i] = (((C[i + 1] * ((m + k - 2) % MOD)) % MOD) *
            (exp(k - 1, MOD - 2) % MOD)) %
           MOD;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i, k = n; j > 0; j--, k--) {
      T[i] += C[k] * A[j];
      T[i] %= MOD;
    }
  }
  for (int i = 1; i < n; i++) cout << T[i] << " ";
  cout << T[n] << "\n";
  return 0;
}
