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 unsigned long long MOD = 1e9 + 7;
unsigned long long quick_pow(unsigned long long num, unsigned long long p) {
  unsigned long long ans = 1;
  if (p == 0) {
    return 1ull;
  }
  if (p == 1ull) {
    return num;
  }
  if (p & 1ull) {
    ans = num;
    p--;
  }
  unsigned long long res = quick_pow(num, p >> 1);
  ans = ans * res % MOD * res % MOD;
  return ans;
}
inline unsigned long long inv(unsigned long long num) {
  return quick_pow(num, MOD - 2);
}
inline unsigned long long C(unsigned long long n, unsigned long long m) {
  unsigned long long ans = 1;
  if (n < m) {
    return 1;
  }
  if (m > n - m) {
    m = n - m;
  }
  for (int i = 1; i <= m; i++) {
    ans = ans * inv(i) % MOD * (n - i + 1) % MOD;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  unsigned long long n, k;
  cin >> n >> k;
  unsigned long long ans = 0, a[2048], b[2048];
  b[0] = 1;
  for (unsigned long long i = 0; i < n; i++) {
    cin >> a[i];
    ans = 0;
    b[i + 1] = b[i] * (k + i) % MOD * inv(i + 1) % MOD;
    for (int j = 0; j <= i; j++) {
      ans = ans + a[j] * b[i - j] % MOD;
    }
    cout << ans % MOD << " ";
  }
  cout << endl;
  return 0;
}
