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;
long long ar[2003];
long long Pow(long long a, long long b) {
  long long c = a, ans = 1;
  for (long long i = 1; i <= b; i <<= 1) {
    if (b & i) ans = (ans * c) % 1000000007;
    c = (c * c) % 1000000007;
  }
  return ans;
}
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%I64d", &ar[i]);
  for (long long i = 1; i <= n; i++) {
    long long sum = ar[i], t = 1;
    for (long long j = i - 1, x = 1, y = k; j >= 1; j--, x++, y++) {
      t = ((t * y) % 1000000007 * Pow(x, 1000000007 - 2)) % 1000000007;
      sum = (sum + (t * ar[j]) % 1000000007) % 1000000007;
    }
    printf("%I64d ", sum);
  }
  printf("\n");
  return 0;
}
