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 = (int)1e9;
const long long INF64 = (long long)1e18;
const long double eps = 1e-9;
const long double pi = 3.14159265358979323846;
const long long MOD = 1000000007;
vector<long long> operator*(const vector<long long> &a,
                            const vector<long long> &b) {
  int n = (int)(a).size();
  vector<long long> ret(n, 0);
  for (int i = 0; i < int(n); ++i) {
    for (int j = 0; j < int(n - i); ++j) {
      ret[i + j] = (ret[i + j] + a[i] * b[j]) % MOD;
    }
  }
  return ret;
}
vector<long long> binpow(const vector<long long> a, int k) {
  vector<long long> e((int)(a).size(), 0);
  e[0] = 1;
  if (k == 0) return e;
  if (k & 1) return a * binpow(a, k - 1);
  vector<long long> b = binpow(a, k / 2);
  return b * b;
}
int main() {
  int n, k;
  cin >> n >> k;
  if (k == 0) {
    for (int i = 0; i < int(n); ++i) {
      int val;
      cin >> val;
      cout << val << ' ';
    }
    return 0;
  }
  vector<long long> a(n);
  for (int i = 0; i < int(n); ++i) cin >> a[i];
  vector<long long> c(n, 1);
  vector<long long> kf = binpow(c, k);
  vector<long long> ans;
  ans = a * kf;
  for (int i = 0; i < int(n); ++i) cout << ans[i] << ' ';
  return 0;
}
