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 Inv(long long x) {
  if (x == 1) return 1;
  return Inv(1000000007 % x) * (1000000007 - 1000000007 / x) % 1000000007;
}
long long inv[2010];
int n, k;
long long c[2010];
void init() {
  for (int i = 1; i <= n; i++) inv[i] = Inv(i);
  for (int i = 0; i <= n; i++) {
    long long up = k + i - 1;
    int down = i;
    c[i] = 1;
    while (down > 0) {
      c[i] = (c[i] * up) % 1000000007;
      c[i] = (c[i] * inv[down]) % 1000000007;
      up--;
      down--;
    }
  }
}
long long s[2010];
long long sum[2010];
int main() {
  scanf("%d %d", &n, &k);
  init();
  for (int i = 0; i < n; i++) scanf("%I64d", &s[i]);
  for (int i = 0; i < n; i++) {
    sum[i] = 0;
    for (int j = 0; j <= i; j++)
      sum[i] = (sum[i] + (s[j] * c[i - j]) % 1000000007) % 1000000007;
  }
  for (int i = 0; i < n; i++) {
    printf("%I64d", sum[i]);
    if (i == n - 1)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}
