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;
int c[2010];
const int mod = (int)1e9 + 7;
inline int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline int mul(int a, int b) { return ((long long)a * (long long)b) % mod; }
int in[2010];
inline int rev(long long a, long long b) {
  if (a == 1) return 1;
  return (((1 - rev(b % a, a) * b) / a) % b + b) % b;
}
int out[2010];
int main() {
  int n, k, i, j;
  scanf("%d%d", &n, &k);
  for (i = 0; i < n; i++) scanf("%d", &in[i]);
  if (k == 0) {
    for (i = 0; i < n; i++) out[i] = in[i];
  } else {
    k--;
    c[0] = 1;
    for (i = 1; i < n; i++) {
      c[i] = mul(mul(c[i - 1], add(k, i)), rev(i, mod));
    }
    for (i = 0; i < n; i++) {
      out[i] = 0;
      for (j = 0; j <= i; j++) out[i] = add(out[i], mul(in[j], c[i - j]));
    }
  }
  for (i = 0; i < n; i++) {
    printf("%d%c", out[i], (i == n - 1 ? '\n' : ' '));
  }
}
