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 MAX = 2005;
const int MOD = 1000000007;
int n, k;
int a[MAX];
int power[MAX];
int c[MAX];
inline int Pow(int x, int k) {
  int res = 1;
  while (k) {
    if (k & 1) res = ((long long)res * x) % MOD;
    x = ((long long)x * x) % MOD;
    k >>= 1;
  }
  return res;
}
inline int Calc(int n, int m) {
  int res = 1;
  for (int i = n; i >= n - m + 1; i--) res = ((long long)res * i) % MOD;
  for (int i = m; i >= 1; i--) res = ((long long)res * power[i]) % MOD;
  return res;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (k == 0) {
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    return 0;
  }
  for (int i = 1; i <= n; i++) power[i] = Pow(i, MOD - 2);
  for (int i = 1; i <= n; i++) c[i] = Calc(k + i - 2, i - 1);
  for (int i = 1; i <= n; i++) {
    int ans = 0;
    for (int j = 1; j <= i; j++)
      ans = (ans + (long long)a[j] * c[i - j + 1]) % MOD;
    printf("%d ", ans);
  }
  return 0;
}
