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 int MAXN = 2010;
const long long MOD = 1E9 + 7;
int N;
long long K;
long long ar[MAXN];
long long C[MAXN];
long long res[MAXN];
long long mpow(long long a, int e) {
  if (e == 0) return 1;
  long long sqrt = mpow(a, e / 2);
  long long res = (sqrt * sqrt) % MOD;
  if (e & 1) {
    res = (res * a) % MOD;
  }
  return res;
}
long long inv(long long a) { return mpow(a, MOD - 2); }
void binom() {
  for (int i = 0; i < N; ++i) {
    long long num = 1;
    long long denom = 1;
    for (int k = 0; k < i; ++k) {
      num *= (K + i - k - 1);
      num %= MOD;
      denom *= (k + 1);
      denom %= MOD;
    }
    C[i] = (num * inv(denom)) % MOD;
  }
}
int main() {
  cin >> N >> K;
  for (int i = 0; i < N; ++i) {
    cin >> ar[i];
  }
  binom();
  for (int i = 0; i < N; ++i) {
    for (int k = i; k < N; ++k) {
      res[k] += ((ar[i] * C[k - i]) % MOD);
      res[k] %= MOD;
    }
  }
  for (int i = 0; i < N; ++i) {
    cout << res[i] << ' ';
  }
  return 0;
}
