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 long long mod = 1000000007;
long long T[2002];
long long A[2002], Ans[2002];
long long n, k;
long long fact(long long i, long long j) {
  long long ret = 1;
  for (; i > j; i--) {
    ret *= i;
    ret %= mod;
  }
  return ret;
}
long long pow(long long n, long long x) {
  if (!x) return 1;
  if (x % 2) return (n * pow(n, x - 1)) % mod;
  long long r = pow(n, x / 2);
  return (r * r) % mod;
}
long long inverse(long long n) { return pow(n, mod - 2); }
long long nchoosek(long long n, long long k) {
  long long a = k, b = n - k;
  if (a < b) swap(a, b);
  n = fact(n, a);
  b = fact(b, 0);
  return (n * inverse(b)) % mod;
}
void preprocess(long long n, long long k) {
  for (long long i = 0; i <= n; i++) T[i] = nchoosek(k + i - 1, i);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> A[i];
  if (k == 0) {
    cout << A[0];
    for (int i = 1; i < n; i++) cout << " " << A[i];
    cout << "\n";
    return 0;
  }
  preprocess(n, k);
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= i; j++) {
      Ans[i] += (A[j] * T[i - j]);
      Ans[i] %= mod;
    }
  cout << Ans[0];
  for (int i = 1; i < n; i++) cout << " " << Ans[i];
  cout << "\n";
  return 0;
}
