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 = 4004;
const long long MOD = 1000000007;
long long tmp[MAX];
long long a[MAX];
long long ans[MAX];
long long m[MAX];
int N;
void mult(long long *a, long long *b) {
  memset(tmp, 0, N * sizeof(long long));
  for (int i = 0; i < N; ++i)
    for (int j = 0; j <= i; ++j) {
      tmp[i] += a[i - j] * b[j];
      tmp[i] %= MOD;
    }
  memcpy(a, tmp, N * sizeof(long long));
}
void power(long long b) {
  ans[0] = 1;
  for (int i = 0; i < N; ++i) m[i] = 1;
  for (; b; b >>= 1) {
    if (b & 1) mult(ans, m);
    mult(m, m);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int K;
  cin >> N >> K;
  for (int i = 0; i < N; ++i) cin >> a[i];
  power(K);
  for (int i = 0; i < N; ++i) {
    if (i) cout << ' ';
    long long ret = 0;
    for (int j = 0; j < N; ++j) {
      ret += ans[i - j] * a[j];
      ret %= MOD;
    }
    cout << ret;
  }
  cout << endl;
  return 0;
}
