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 = 1000000007;
long long a[2005], b[2005], c[2005];
void mult(long long a[2005], long long b[2005], int n) {
  int i, j;
  for (i = 0; i < n; i++) {
    c[i] = 0;
    for (j = 0; j < n; j++) {
      c[i] += a[j] * (j > i ? 0 : b[i - j]);
      c[i] %= MOD;
    }
  }
  for (i = 0; i < n; i++) a[i] = c[i];
}
void show(long long b[][2005], int k) {
  printf("b^%d:\n", k);
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) {
      printf("%d ", b[i][j]);
    }
    putchar('\n');
  }
}
int main() {
  int n, k;
  while (scanf("%d%d", &n, &k) != EOF) {
    for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
    for (int i = 0; i < n; i++) b[i] = 1;
    while (k) {
      if (k & 1) mult(a, b, n);
      mult(b, b, n);
      k >>= 1;
    }
    for (int i = 0; i < n; i++)
      printf("%I64d%c", a[i], (i == n - 1) ? '\n' : ' ');
  }
  return 0;
}
