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;
using namespace std;
long long mod = 1e9 + 7;
long long quick_mod(long long a, long long b) {
  long long ans = 1;
  a %= mod;
  while (b) {
    if (b % 2 != 0) {
      ans = ans * a % mod;
      b--;
    }
    b /= 2;
    a = a * a % mod;
  }
  return ans;
}
long long judge(long long n, long long m) {
  if (m > n) return 0;
  long long ans = 1;
  for (long long i = 1; i <= m; i++) {
    long long a = (n + i - m) % mod;
    long long b = i % mod;
    ans = ans * (a * quick_mod(b, mod - 2) % mod) % mod;
  }
  return ans;
}
long long work(long long n, long long m) {
  if (m == 0) return 1;
  return judge(n % mod, m % mod) * work(n / mod, m / mod) % mod;
}
const int MAXN = 2e3 + 5;
long long ans[MAXN], a[MAXN];
long long sum[MAXN];
long long C[MAXN << 1];
int main() {
  long long m, n, i, j, k;
  ios::sync_with_stdio(false);
  ;
  cin >> m >> k;
  for (i = 1; i <= m; i++) cin >> a[i];
  long long c1 = k - 1, c2 = 0;
  for (i = 0; i <= m; i++) {
    C[i] = work(c1, c2);
    c1++, c2++;
  }
  for (i = 1; i <= m; i++) {
    long long c = 0;
    for (j = i; j <= m; j++) {
      ans[j] = (ans[j] + C[c++] * a[i]) % mod;
    }
  }
  for (i = 1; i <= m; i++) cout << ans[i] << ' ';
  cout << endl;
}
