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 n, k;
int a[2014];
int C[2014];
const int mod = 1e9 + 7;
int binpow(long long a, int x) {
  long long res = 1;
  while (x) {
    if (x & 1) res = res * a % mod;
    a = a * a % mod;
    x >>= 1;
  }
  return res;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> a[i];
  C[0] = 1;
  for (int i = 1; i < 2014; ++i) {
    C[i] = C[i - 1] * 1ll * (k + i - 1) % mod;
    C[i] = C[i] * 1ll * binpow(i, mod - 2) % mod;
  }
  for (int i = 0; i < n; ++i) {
    long long res = 0;
    for (int j = i; j > -1; --j) {
      res = res + a[j] * 1ll * (C[i - j]);
      res %= mod;
    }
    cout << res << ' ';
  }
  return 0;
}
