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 maxn = 2000 + 5;
const int MOD = 1000000000 + 7;
int n, k;
int a[maxn], e[maxn], s[maxn], t[maxn];
inline void mm(int a[], int b[]) {
  for (int i = 0; i < n; i++) {
    t[i] = 0;
    for (int j = 0; j <= i; j++)
      t[i] = ((long long)a[j] * b[i - j] + t[i]) % MOD;
  }
  for (int i = 0; i < n; i++) a[i] = t[i];
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    e[i] = 1;
  }
  s[0] = 1;
  while (k) {
    if (k & 1) mm(s, e);
    mm(e, e);
    k >>= 1;
  }
  mm(a, s);
  for (int i = 0; i < n; i++) printf("%d ", a[i]);
  return 0;
}
