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 long long p = 1000000007;
long long Div(long long a, long long b);
long long Pow(long long a, long long b);
long long Div(long long a, long long b) {
  if (b % p == 0) return -1;
  return (a * Pow(b, p - 2)) % p;
}
long long Pow(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = Pow(a, b / 2);
  if (b % 2 == 0)
    return (ans * ans) % p;
  else
    return (((ans * ans) % p) * a) % p;
}
int main() {
  long long n, k, a[3000], C[3000], i, s[3000], j;
  scanf("%I64d%I64d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (k == 0) {
    for (i = 1; i <= n; i++) printf("%I64d ", a[i]);
    return 0;
  }
  k--;
  C[1] = 1;
  for (i = 2; i <= n; i++) C[i] = (Div(k + i - 1, i - 1) * C[i - 1]) % p;
  for (i = 1; i <= n; i++) {
    s[i] = 0;
    for (j = i; j >= 1; j--) s[i] = (s[i] + a[j] * C[i - j + 1]) % p;
  }
  for (i = 1; i <= n; i++) printf("%I64d ", s[i]);
  return 0;
}
