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 double eps = 1e-9;
const double pi = acos(-1.0);
const long long inf = 1e18;
const long long mod = 1000000007;
using namespace std;
long long a[2222], ans[2222];
int n, k;
long long power(long long a, long long pow) {
  if (pow == 0) return 1;
  if (pow % 2 == 1) {
    long long cur = power(a, pow - 1);
    cur *= a;
    cur %= mod;
    return cur;
  }
  long long cur = power(a, pow / 2);
  cur *= cur;
  cur %= mod;
  return cur;
}
long long inv(long long a) { return power(a, mod - 2); }
int main(int argc, const char* argv[]) {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  if (k == 0) {
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    return 0;
  }
  --k;
  for (int i = 1; i <= n; i++) {
    long long cur = 1, diff = 0;
    for (int j = i; j <= n; j++) {
      ans[j] += (cur * a[i]) % mod;
      ans[j] %= mod;
      diff++;
      cur *= diff + k;
      cur %= mod;
      cur *= inv(diff);
      cur %= mod;
    }
  }
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
