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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int a[200005], b[200005], c[200005];
long long int bigmod(long long int n, long long int p) {
  if (p == 0) return 1;
  long long int x = bigmod(n, p / 2);
  x = (x * x) % 1000000007;
  if (p % 2 == 1) x = (x * n) % 1000000007;
  return x;
}
int main() {
  long long int n, i, j, k;
  while (cin >> n >> k) {
    for (long long int i = 1; i < n + 1; i++) {
      cin >> a[i];
      b[i] = a[i];
    }
    c[0] = 1;
    for (i = 1; i <= n; i++) {
      c[i] = c[i - 1] * (k + i - 1);
      c[i] %= 1000000007;
      c[i] = c[i] * bigmod(i, 1000000007 - 2);
      c[i] %= 1000000007;
    }
    for (i = 1; i <= n; i++) {
      for (j = i + 1; j <= n; j++) {
        b[j] = b[j] + (a[i] * c[j - i]) % 1000000007;
        b[j] %= 1000000007;
      }
    }
    for (i = 1; i <= n; i++) cout << b[i] << " ";
    cout << endl;
  }
  return 0;
}
