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>
#pragma GCC optimize("O2")
using namespace std;
const long long N = 2010;
const long long mod = 1e9 + 7;
long long g[N];
long long n, k;
long long a[N];
long long temp[N];
long long b[N];
bool flag;
void mul(long long a[], long long b[], long long c[]) {
  memset(temp, 0, sizeof temp);
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= i; j++)
      temp[i] = (temp[i] + a[j] * b[i - j + 1] % mod) % mod;
  for (long long i = 1; i <= n; i++) c[i] = temp[i];
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n >> k;
  g[1] = 1;
  for (long long i = 1; i <= n; i++) b[i] = 1;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  while (k) {
    if (k & 1) mul(g, b, g);
    k >>= 1;
    mul(b, b, b);
  }
  for (long long i = 1; i <= n; i++) {
    long long x = 0;
    for (long long j = 1; j <= i; j++) x = (x + g[i - j + 1] * a[j]) % mod;
    cout << x << " ";
  }
  cout << endl;
  return 0;
}
