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 long long mod = 1e9 + 7;
const int MAX = 2001;
long long invk[2005];
long long Fn[4005];
long long exp(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = (res * a) % mod;
    }
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long f(long long fn, long long in) {
  long long res = 1;
  for (; in <= fn; in++) res = (res * in) % mod;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  invk[0] = 1ll;
  for (long long n = 1; n < MAX; n++) invk[n] = invk[n - 1] * n % mod;
  for (long long n = 0; n < MAX; n++) invk[n] = exp(invk[n], mod - 2);
  int n;
  long long k;
  cin >> n >> k;
  long long ar[n + 5];
  for (int i = 1; i <= n; i++) cin >> ar[i];
  if (!k) {
    for (int i = 1; i <= n; i++) cout << ar[i] << " ";
    return 0;
  }
  Fn[0] = k;
  for (long long i = 1; i <= n + 7; i++) Fn[i] = Fn[i - 1] * (i + k) % mod;
  for (long long i = 1; i <= n; i++) {
    long long cur = 0;
    for (long long j = i; j >= 1; j--) {
      long long d = i - j;
      long long cnt;
      if (!d)
        cnt = 1ll;
      else
        cnt = Fn[d - 1];
      cur += (ar[j] * (cnt * invk[d] % mod) % mod);
      cur %= mod;
    }
    cout << cur << " ";
  }
}
