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 N = 2050, mod = 1e9 + 7;
int n, k;
vector<long long> a(N), base(N);
void print() {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
}
vector<long long> mul(vector<long long> &a, vector<long long> &b) {
  vector<long long> res(N);
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++) res[i] = (res[i] + a[j] * b[i - j]) % mod;
  return res;
}
vector<long long> pw(int p) {
  if (p == 1) return base;
  vector<long long> ret = pw(p / 2);
  ret = mul(ret, ret);
  if (p & 1) ret = mul(ret, base);
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i], base[i] = 1;
  if (!k) return print(), 0;
  vector<long long> p = pw(k);
  a = mul(a, p);
  print();
}
