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;
const long long mod = 1000000007;
int n, k;
long long a[2001], c[2001];
long long lpow(long long a, int e) {
  if (e == 0) return 1;
  if (e % 2) {
    long long ret = lpow(a, e / 2);
    return ((ret * ret) % mod * a) % mod;
  } else {
    long long ret = lpow(a, e / 2);
    return (ret * ret) % mod;
  }
}
long long solve(long long a, long long b) {
  long long ret = 1;
  for (int i = 1; i <= b; ++i) {
    ret = (ret * (a - b + i)) % mod;
    ret = (ret * lpow(i, mod - 2)) % mod;
  }
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  if (k == 0) {
    for (int i = 0; i < n; ++i) cout << a[i] << " ";
    return 0;
  }
  for (int i = 0; i < n; ++i) {
    c[i] = solve(k - 1 + i, i);
  }
  for (int i = 0; i < n; ++i) {
    long long ans = 0;
    for (int j = 0; j <= i; ++j) ans = (ans + (c[j] * a[i - j]) % mod) % mod;
    cout << ans << " ";
  }
  return 0;
}
