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 double eps = 1e-8;
const int maxn = 100005;
const long long mod = 1e9 + 7;
vector<long long> a, b;
int n, k;
vector<long long> calc(vector<long long>& s, vector<long long>& t) {
  vector<long long> tmp;
  tmp.resize(n + 1, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= i; j++) {
      tmp[i] += s[i - j] * t[j];
      tmp[i] %= mod;
    }
  }
  return tmp;
}
void pow_mod(long long p) {
  while (p) {
    if (p & 1) {
      a = calc(a, b);
    }
    b = calc(b, b);
    p >>= 1;
  }
}
int main() {
  scanf("%d %d", &n, &k);
  a.resize(n + 1), b.resize(n + 1);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = 1;
  pow_mod(k);
  for (int i = 0; i < n; i++) cout << a[i] << ' ';
  cout << endl;
  return 0;
}
