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, MOD = 1e9 + 7;
int n, K, a[maxn + 5], c[maxn + 5], INV[maxn + 5], ans[maxn + 5];
inline int C(int x, int y) {
  int ans = INV[y];
  for (int i = 1; i <= y; i++) ans = (long long)ans * (x - i + 1) % MOD;
  return ans;
}
inline void Make(int n) {
  INV[0] = INV[1] = 1;
  for (int i = 2; i <= n; i++)
    INV[i] = (long long)(MOD - MOD / i) * INV[MOD % i] % MOD;
  for (int i = 1; i <= n; i++) INV[i] = (long long)INV[i - 1] * INV[i] % MOD;
  for (int i = 0; i <= n; i++) c[i] = C(i + K - 1, i);
}
inline void AMOD(int &x, int tem) {
  if ((x += tem) >= MOD) x -= MOD;
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (!K) {
    for (int i = 1; i <= n; i++)
      i < n ? printf("%d ", a[i]) : printf("%d\n", a[i]);
    return 0;
  }
  Make(n);
  for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++) AMOD(ans[j], (long long)a[i] * c[j - i] % MOD);
  for (int i = 1; i <= n; i++)
    i < n ? printf("%d ", ans[i]) : printf("%d\n", ans[i]);
  return 0;
}
