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;
static const int INF = 500000000;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
int n, k;
int a[2005];
long long int C[2005];
const long long int mod = 1000000007;
long long int extgcd(long long int a, long long int b, long long int& x,
                     long long int& y) {
  long long int d = a;
  if (b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}
long long int invmod(long long int a) {
  long long int x, y;
  extgcd(a, mod, x, y);
  return (x % mod + mod) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  if (k == 0) {
    for (int i = 0; i < n; ++i) printf("%d%c", a[i], i == n - 1 ? '\n' : ' ');
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    C[i] = 1;
    for (int j = 0; j < i; ++j)
      C[i] = (C[i] * (i + k - 1 - j) % mod * invmod(j + 1) % mod);
  }
  for (int i = 0; i < n; ++i) {
    long long int sum = 0;
    for (int j = 0; j < i + 1; ++j) {
      sum += C[i - j] * a[j];
      sum %= mod;
    }
    cout << sum;
    printf("%c", i == n - 1 ? '\n' : ' ');
  }
  return 0;
}
