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;
inline long long in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline string getStr() {
  char ch[200000];
  scanf("%s", ch);
  return ch;
}
inline char getCh() {
  char ch;
  scanf(" %c", &ch);
  return ch;
}
template <class P, class Q>
inline P smin(P &a, Q b) {
  if (b < a) a = b;
  return a;
}
template <class P, class Q>
inline P smax(P &a, Q b) {
  if (a < b) a = b;
  return a;
}
const long long MOD = 1e9 + 7;
const long long MAX_N = 2010;
const long long MA_LG = 21;
const long long base = 29;
long long a[MAX_N];
long long s[MAX_N];
long long C[MAX_N];
long long fact[MAX_N];
inline long long power(long long a, long long b) {
  if (!b) return 1;
  long long res = power(1LL * a * a % MOD, b / 2);
  if (b % 2)
    return 1LL * a * res % MOD;
  else
    return 1LL * res % MOD;
}
inline void rel(long long &x) { x %= MOD; }
inline long long CalcC(long long n, long long r) {
  long long res = power(r, MOD - 2);
  while (r--) res *= n, n--;
  return res;
}
int32_t main() {
  long long n = in(), k = in();
  for (long long i = 0; i < n; i++) a[i] = in();
  fact[0] = C[0] = 1;
  for (long long i = 1; i < MAX_N; i++) fact[i] = 1LL * fact[i - 1] * i % MOD;
  for (long long i = 1; i < MAX_N; i++) C[i] = CalcC(i + k - 1, i);
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j <= i; j++) {
      s[i] += 1LL * C[i - j] * a[j] % MOD, rel(s[i]);
    }
    cout << s[i] << " ";
  }
  cout << endl;
}
