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 MOD = 1e9 + 7;
const int MAXN = 2e4 + 5;
const int N = 1e4 + 10;
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    n >>= 1;
  }
  return res;
}
long long F[MAXN], ans[MAXN];
int n, k, a[MAXN];
void Init() {
  F[0] = 1;
  for (int i = 1; i <= n; ++i) {
    F[i] = (F[i - 1] * (k + i - 1)) % MOD;
    F[i] = (F[i] * power(i, MOD - 2)) % MOD;
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  Init();
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= i; ++j)
      ans[i] = (ans[i] + (a[j] * F[i - j]) % MOD) % MOD;
  }
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  return 0;
}
