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 inf = 1e9 + 5;
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
const int MAXN = 2345;
const long long mod = 1e9 + 7;
int N, K;
int ar[MAXN];
long long res[MAXN];
long long C[MAXN];
long long F[MAXN];
inline void add(long long &a, long long b) {
  a = (a + b);
  if (a >= mod) a -= mod;
}
long long pw(long long a, long long b) {
  long long res = 1LL;
  for (; b; b >>= 1, a = (a * a) % mod)
    if (b & 1) res = (res * a) % mod;
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> K;
  ++K;
  for (int(i) = (1); (i) <= (N); (i)++) cin >> ar[i];
  if (K == 1) {
    for (int(i) = (1); (i) <= (N); (i)++) cout << ar[i] << ' ';
    return 0;
  }
  F[0] = 1LL;
  for (int(i) = (1); (i) <= (2222); (i)++) F[i] = (F[i - 1] * i) % mod;
  int go = K - 2;
  for (int(i) = (K - 2); (i) <= (K + 2111); (i)++) {
    C[i - go] = 1LL;
    for (int(j) = (K - 1); (j) <= (i); (j)++) C[i - go] = (C[i - go] * j) % mod;
    C[i - go] = (C[i - go] * pw(F[i - (K - 2)], mod - 2)) % mod;
  }
  for (int(i) = (1); (i) <= (N); (i)++) {
    long long res = 0LL;
    for (int(j) = (1); (j) <= (i); (j)++)
      add(res, (C[K - 2 + i - j - go] * ar[j]) % mod);
    cout << res << ' ';
  }
  return 0;
}
