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 C[2222], Ans[2222];
int A[2222];
int N, K;
const long long P = 1000000000LL + 7;
long long fast_mod(int x, int y) {
  if (y == 0) {
    return 1LL;
  }
  long long ret = fast_mod(x, y >> 1);
  ret *= ret;
  ret %= P;
  if (y & 1) {
    ret *= x;
    ret %= P;
  }
  return ret;
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
  }
  C[0] = 1;
  for (int i = 1; i < N; i++) {
    C[i] = C[i - 1] * (K + i - 1);
    C[i] %= P;
    C[i] *= fast_mod(i, P - 2);
    C[i] %= P;
  }
  for (int i = N - 1; i >= 0; i--) {
    for (int j = i; j >= 0; j--) {
      Ans[i] += A[i - j] * C[j];
      Ans[i] %= P;
    }
  }
  for (int i = 0; i < N; i++) {
    printf("%d", Ans[i]);
    printf(i == N - 1 ? "\n" : " ");
  }
  return 0;
}
