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 = 1000000000 + 7;
const int N = 2000;
int n, m, a[N], b[N], inverse[N + 1];
int main() {
  scanf("%d%d", &n, &m);
  inverse[1] = 1;
  for (int i = 2; i <= n; ++i) {
    inverse[i] = (long long)(MOD - MOD / i) * inverse[MOD % i] % MOD;
  }
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  memset(b, 0, sizeof(b));
  if (m == 0) {
    for (int i = 0; i < n; ++i) {
      b[i] = a[i];
    }
  } else {
    m--;
    for (int i = 0; i < n; ++i) {
      long long coefficient = 1;
      for (int j = 0; i + j < n; ++j) {
        if (j > 0) {
          coefficient = coefficient * (m + j) % MOD;
          coefficient = coefficient * inverse[j] % MOD;
        }
        b[i + j] += coefficient * a[i] % MOD;
        b[i + j] %= MOD;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    printf("%d%c", b[i], i == n - 1 ? '\n' : ' ');
  }
  return 0;
}
