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;
long long FastPower(long long x, long long y);
long long InverseMod(long long b);
const long long Mod = 1e9 + 7;
long long a[2020];
int main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) scanf("%I64d", &a[i]);
  for (long long i = 0; i < n; i++) {
    long long Res = 0;
    for (long long j = i, p = 1, d = 1, r = 1; j >= 0;
         j--, p = (p * (k + r - 1)) % Mod, d = (d * r) % Mod, r++)
      Res = (Res + ((p * InverseMod(d)) % Mod) * a[j]) % Mod;
    printf("%I64d ", Res);
  }
}
long long FastPower(long long x, long long y) {
  if (!y) return 1;
  long long Res = FastPower(x, y >> 1);
  Res = (Res * Res) % Mod;
  if (y & 1) Res *= x;
  return Res % Mod;
}
long long InverseMod(long long b) { return FastPower(b, Mod - 2) % Mod; }
