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;
const int mod = 1e9 + 7;
int n, K, a[N], c[N], jc[N], cc[N];
long long ni(long long x) {
  long long s = 1;
  int y = mod - 2;
  x %= mod;
  while (y) {
    if (y & 1) s = s * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return s;
}
void init() {
  int i;
  for (i = 1, jc[0] = 1; i <= n; i++) jc[i] = (long long)jc[i - 1] * i % mod;
  cc[0] = 1;
  for (i = K + 1; i <= K + n; i++)
    cc[i - K] = (long long)cc[i - K - 1] * i % mod;
  for (i = 0; i <= n; i++) c[i] = (long long)cc[i] * ni(jc[i]) % mod;
}
void iWork() {
  cin >> n >> K;
  K--;
  int i, j;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (K == -1) {
    for (i = 1; i <= n; i++) printf("%d\n", a[i]);
    return;
  }
  init();
  for (i = 1; i <= n; i++) {
    long long ans = 0;
    for (j = 1; j <= i; j++) {
      (ans += (long long)a[j] * c[i - j] % mod) %= mod;
    }
    printf("%d ", ans);
  }
}
int main() {
  iWork();
  return 0;
}
