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;
int m;
const long long mod = 1000000007;
void mul(vector<int> &a, vector<int> &b) {
  vector<int> d(m + 1);
  d[0] = 0;
  int i, j, k;
  for (int i = 1; i <= m; i++) {
    long long s = 0;
    for (j = i, k = 1; j >= 1; j--, k++) {
      int x = a[j], y = b[k];
      s += ((long long)x * (long long)y) % mod, s %= mod;
    }
    d[i] = s;
  }
  for (i = 1; i <= m; i++) a[i] = d[i];
}
void pow(vector<int> &A, long long p) {
  vector<int> c(m + 1);
  for (int i = 1; i <= m; i++) c[i] = 1;
  p--;
  while (p > 0) {
    if (p & 1) mul(c, A);
    mul(A, A);
    p >>= 1;
  }
  for (int i = 1; i <= m; i++) A[i] = c[i];
}
int a[2005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> m >> n;
  for (int i = 1; i <= m; i++) cin >> a[i];
  int i, j, k;
  if (n == 0) {
    for (int i = 1; i <= m; i++) cout << a[i] << " ";
    return 0;
  }
  vector<int> c(m + 1);
  for (int i = 1; i <= m; i++) c[i] = 1;
  pow(c, n);
  for (i = 1; i <= m; i++) {
    long long s = 0;
    for (j = i, k = 1; j >= 1; j--, k++) {
      int x = c[j], y = a[k];
      s += ((long long)x * (long long)y) % mod, s %= mod;
    }
    cout << s << " ";
  }
  return 0;
}
