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 = 1000000007LL;
const int LIM = 2009;
int n;
long long k;
long long a[LIM];
long long ans;
long long Ck[LIM];
void addM(long long &A, long long B) { A = (A + B) % mod; }
void subM(long long &A, long long B) {
  A = (A - B) % mod;
  if (A < 0) A += mod;
}
void mulM(long long &A, long long B) { A = (A * B) % mod; }
long long powM(long long bs, long long pwr) {
  long long func = 1LL;
  int bfst = 64 - __builtin_clzll(pwr);
  for (int i = bfst - 1; i >= 0; i--) {
    func = (func * func) % mod;
    if (pwr & (1LL << i)) func = (func * bs) % mod;
  }
  return func;
}
void init() {
  Ck[0] = 1;
  for (int i = 1; i < LIM; i++)
    Ck[i] = (((Ck[i - 1] * (k + (long long)i)) % mod) *
             powM((long long)i, mod - 2LL)) %
            mod;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (k == 0) cout << a[i] << " ";
  }
  if (k == 0) {
    cout << "\n";
    return 0;
  } else
    k--;
  init();
  for (int i = 1; i <= n; i++) {
    ans = 0LL;
    for (int j = 1; j <= i; j++) {
      addM(ans, (Ck[j - 1] * a[i - j + 1]) % mod);
    }
    cout << ans << " ";
  }
  cout << "\n";
}
