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;
long long N, K, a[2005], ans[2005];
long long POW(long long x, int n) {
  if (n == 0) return 1;
  if (n == 1) return x;
  long long h = POW(x, n / 2);
  h = (h * h) % 1000000007;
  if (n % 2) h = (h * x) % 1000000007;
  return h;
}
int main() {
  int i, j;
  long long x;
  while (~scanf("%d%d", &N, &K)) {
    a[0] = 0;
    for (i = 1; i <= N; i++) scanf("%d", &a[i]);
    if (K == 0) {
      for (i = 1; i <= N; i++) printf("%d ", a[i]);
      goto A;
    }
    if (K == 1) {
      long long x = 0;
      for (i = 1; i <= N; i++) x += a[i], printf("%d ", x);
      goto A;
    }
    for (i = 1; i <= N; i++) {
      x = (a[i] + K * a[i - 1]) % 1000000007;
      for (j = 1; j < i - 1; j++)
        x = (x +
             (POW(3, (i - j - 1)) * (K - 1)) % 1000000007 * a[j] % 1000000007) %
            1000000007;
      printf("%d ", x);
    }
  A:
    printf("\n");
  }
  return 0;
}
