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 INF = 2147483647;
const long long LLINF = 9223372036854775807LL;
const int mod = 1000000007;
const int maxn = 2010;
int powmod(int x, int p) {
  if (p == 0) return 1;
  if (p & 1)
    return (long long)x * powmod(x, p - 1) % mod;
  else {
    long long t = powmod(x, p >> 1);
    return t * t % mod;
  }
}
int koeff[maxn] = {};
int ans[maxn] = {};
int a[maxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  koeff[0] = 1;
  for (int i = 1; i < maxn; ++i) {
    koeff[i] = ((koeff[i - 1] * (long long)(k + i - 1)) % mod *
                (long long)powmod(i, mod - 2)) %
               mod;
  }
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j <= i; ++j) {
      ans[i] = (ans[i] + (long long)koeff[i - j] * a[j]) % mod;
    }
  }
  for (int i = 0; i < n; ++i) printf("%d ", ans[i]);
  return 0;
}
