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 = 1000000007;
const long long inf = LLONG_MAX;
const long long ninf = LLONG_MIN;
const long double eps = 1e-12;
const long long N = 1000005;
const long long LOGN = 19;
long long n, k;
long long a[N], ans[N], C[N];
long long power(long long a, long long b) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > mod) x %= mod;
    }
    y = (y * y);
    if (y > mod) y %= mod;
    b /= 2;
  }
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    C[i] = C[i - 1] * (k + i - 1) % mod * power(i, mod - 2) % mod;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i + 1; j++) {
      if (k - 1 <= k + i - j) {
        ans[i] = (ans[i] + C[i - j] * a[j]) % mod;
      }
    }
  }
  for (int i = 0; i < n; i++) cout << (k ? ans[i] : a[i]) << " ";
  cout << "\n";
  return 0;
}
