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 long long MOD = 1000000007;
long long co[2200], a[2200];
int n, m;
long long inv(int a, int k) {
  if (k == 0) return 1;
  if (k == 1) return a;
  long long tmp = inv(a, k / 2);
  if (k & 1)
    return ((tmp * tmp) % MOD * a) % MOD;
  else
    return (tmp * tmp) % MOD;
}
int main() {
  scanf("%d%d", &n, &m);
  co[0] = 1;
  for (int i = 1; i < n; ++i)
    co[i] = co[i - 1] * (m + i - 1) % MOD * inv(i, MOD - 2) % MOD;
  for (int i = 0; i < n; ++i) scanf("%I64d", &a[i]);
  for (int i = 0; i < n; ++i) {
    long long ans = 0;
    for (int j = 0; j <= i; ++j) ans = (ans + co[i - j] * a[j]) % MOD;
    printf("%I64d%c", ans, i + 1 == n ? '\n' : ' ');
  }
  return 0;
}
