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 mod = 1e9 + 7;
int qpow(int b, int exp) {
  if (exp == 0)
    return 1;
  else if (exp % 2)
    return (1LL * b * qpow(b, exp - 1)) % mod;
  else {
    int r = qpow(b, exp / 2);
    return (1LL * r * r) % mod;
  }
}
const int MN = 2002;
int N, M, i, arr[MN], coef[MN], j, res[MN];
int main() {
  for (scanf("%d%d", &N, &M), i = 1; i <= N; i++) scanf("%d", &arr[i]);
  if (M == 0) {
    for (i = 1; i <= N; i++) {
      printf("%d ", arr[i]);
    }
    return 0;
  }
  coef[0] = 1;
  for (i = 1; i <= N; i++)
    coef[i] =
        ((1LL * coef[i - 1] * (M + i - 1)) % mod * qpow(i, mod - 2)) % mod;
  for (i = 1; i <= N; i++) {
    for (j = i; j <= N; j++)
      res[j] = (1LL * res[j] + 1LL * arr[i] * coef[j - i]) % mod;
  }
  for (i = 1; i <= N; i++) printf("%d ", res[i]);
  getchar();
  getchar();
  return 0;
}
