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;
int ntest = 0, test = 0;
inline void init();
inline void run();
inline void stop() { ntest = test - 1; }
int main() {
  init();
  while (++test <= ntest) {
    run();
  }
  return 0;
}
const int INF = (int)1E9 + 5;
const double EPS = 1E-11;
const long long MOD = (long long)1E9 + 7;
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
inline void init() { ntest = 1; }
long long mu(long long a, long long n) {
  long long rs = 1;
  for (; n; n >>= 1) {
    if (n & 1) rs = (rs * a) % MOD;
    a = (a * a) % MOD;
  }
  return rs;
}
vector<int> a;
inline void run() {
  int n, k;
  cin >> n >> k;
  a.resize(n);
  for (int i = 0; i < (n); i++) cin >> a[i];
  if (k == 0) {
    for (int i = 0; i < (n); i++) cout << a[i] << " ";
    return;
  }
  for (int i = 0; i < (n); i++) {
    long long total = a[i];
    long long cur = 1;
    for (int j = (1); j <= (i); j++) {
      cur = (cur * (k + j - 1)) % MOD;
      cur = (cur * mu(j, MOD - 2)) % MOD;
      total = (total + (cur * a[i - j]) % MOD) % MOD;
    }
    cout << total << " ";
  }
}
