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 = 2003;
const int MOD = 1000000007;
long long nCr[N];
int Power(int base, int p) {
  long long res = 1;
  long long curr = base;
  for (int i = 0; i < 32; ++i) {
    if ((p >> i) & 1) {
      res = res * curr % MOD;
    }
    curr = curr * curr % MOD;
  }
  return res;
}
void Build(int k) {
  nCr[0] = 1;
  for (int i = k + 1; i < N + k + 1; ++i) {
    nCr[i - k] = nCr[i - k - 1] * Power(i - k, MOD - 2) % MOD;
    nCr[i - k] = nCr[i - k] * i % MOD;
  }
}
int arr[N];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  --k;
  Build(k);
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    int res = 0;
    for (int j = 0; j <= i; ++j) {
      res = (res + 1LL * arr[j] * nCr[i - j]) % MOD;
    }
    cout << res << ' ';
  }
}
