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 A[2010], C[2010][2010], trans[2010][2010], res[2010][2010];
int mul(int a, int b) {
  int x = 1LL * a * b;
  if (x < 1000000007) return x;
  return x % 1000000007;
}
int add(int a, int b) {
  int x = (a + b);
  if (x < 1000000007) return x;
  return x - 1000000007;
}
int n;
void mult(int A[2010][2010], int B[2010][2010]) {
  for (int i = 0; i < n; i++) {
    C[0][i] = 0;
    for (int j = 0; j <= i; j++) {
      C[0][i] = add(C[0][i], mul(B[j][i], A[0][j]));
    }
  }
  for (int i = 1; i < n; i++)
    for (int j = i; j < n; j++) {
      C[i][j] = C[i - 1][j - 1];
      A[i][j] = C[i][j];
    }
  for (int i = 0; i < n; i++) A[0][i] = C[0][i];
}
void expo(int A[2010][2010], int B[2010][2010], int k) {
  for (int i = 0; i < n; i++) B[i][i] = 1;
  while (k) {
    if (k & 1) mult(B, A);
    mult(A, A);
    k /= 2;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> A[i];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) trans[j][i] = 1;
  }
  expo(trans, res, k);
  for (int i = 0; i < n; i++) {
    int ans = 0;
    for (int j = 0; j < n; j++) {
      ans = add(ans, mul(A[j], res[j][i]));
    }
    cout << ans << " ";
  }
  cout << '\n';
}
