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>
const int p = 1000000007;
int f[3000], g[3000], h[3000], v[3000];
int i, j, m, n;
inline int power(int x, int y) {
  if (y == 1) return x;
  int t = power(x, y / 2);
  t = 1LL * t * t % p;
  if (y & 1) t = 1LL * t * x % p;
  return t;
}
inline int C(int x, int y) {
  if ((x < 0) || (y < 0)) return 0;
  int t = 1;
  x = x % p;
  if (x < y) return 0;
  for (int i = x; i > x - y; i--) t = 1LL * t * i % p;
  for (int i = 1; i <= y; i++) t = 1LL * t * v[i] % p;
  return t;
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d", &g[i]);
  if (m == 0) {
    for (i = 1; i <= n; i++) printf("%d ", g[i]);
    return 0;
  }
  for (i = 1; i <= n; i++) v[i] = power(i, p - 2);
  for (i = 1; i <= n; i++) h[i] = C(m - 1 + i - 1, i - 1);
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++) f[j] = (f[j] + 1LL * g[i] * h[j - i + 1]) % p;
  for (i = 1; i <= n; i++) printf("%d ", f[i]);
  return 0;
}
