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 int N = 2010, mod = 1e9 + 7;
int c[N], a[N], s[N], f1[N], t[N];
void mul(int c[], int a[], int b[]) {
  memset(t, 0, sizeof t);
  for (int i = 1; i < N; i++)
    for (int k = 1; k <= i; k++)
      t[i] = (t[i] + (long long)a[k] * b[i - k + 1] % mod) % mod;
  for (int i = 1; i < N; i++) c[i] = t[i];
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> c[i];
  if (k == 0) {
    for (int i = 1; i <= n; i++) cout << c[i] << " ";
    return 0;
  }
  k--;
  for (int i = 1; i < N; i++) f1[i] = 1;
  for (int i = 1; i < N; i++) a[i] = 1;
  while (k) {
    if (k & 1) mul(f1, f1, a);
    mul(a, a, a);
    k >>= 1;
  }
  int t[N];
  memset(t, 0, sizeof t);
  for (int i = 1; i < N; i++)
    for (int k = 1; k <= i; k++)
      t[i] = (t[i] + (long long)c[k] * f1[i - k + 1] % mod) % mod;
  for (int i = 1; i <= n; i++) cout << t[i] << " ";
}
