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 V = 2010;
const long long MOD = 1e9 + 7;
int n, k;
long long data[V][V];
int a[V], inv[V];
long long c[V];
void prepare() {
  inv[1] = 1;
  for (int i = 2; i < V; i++)
    inv[i] = (MOD - MOD / i) * inv[(int)(MOD % i)] % MOD;
}
long long C(long long x, int y) {
  long long ret = 1;
  for (int i = 0; i < y; ++i) {
    ret = ret * (x - i) % MOD * inv[i + 1] % MOD;
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  prepare();
  for (int i = 0; i < n; ++i) c[i] = C(k + i - 1, i);
  for (int i = 0; i < n; ++i) {
    for (int j = i; j < n; ++j)
      data[i][j] = (data[i][j] + c[j - i] * a[i]) % MOD;
  }
  for (int i = 0; i < n; ++i) {
    long long ret = 0;
    for (int j = 0; j < n; ++j) ret = (ret + data[j][i]) % MOD;
    if (i != 0) printf(" ");
    printf("%I64d", ret);
  }
  printf("\n");
}
