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 pm(int x, int e) {
  long long res = 1;
  for (long long b = x; e; b = b * b % MOD, e >>= 1)
    if (e & 1) res = res * b % MOD;
  return res;
}
long long binom(int n, int k) {
  long long res = 1;
  for (int i = 1; i <= k; ++i) {
    res *= n - i + 1;
    res %= MOD;
    res *= pm(i, MOD - 2);
    res %= MOD;
  }
  return res;
}
int main() {
  int N, K;
  while (cin >> N >> K) {
    vector<int> a(N);
    for (int i = 0; i < int(N); ++i) cin >> a[i];
    if (K == 0) {
      for (int i = 0; i < int(N); ++i)
        cout << a[i] << (i == N - 1 ? "\n" : " ");
      continue;
    }
    --K;
    vector<long long> b(N);
    for (int i = 0; i < int(N); ++i) b[i] = binom(K + i, i);
    vector<long long> s(N);
    for (int i = 0; i < int(N); ++i)
      for (int j = 0; j < int(i + 1); ++j)
        s[i] = (s[i] + a[j] * b[i - j]) % MOD;
    for (int i = 0; i < int(N); ++i) cout << s[i] << (i == N - 1 ? "\n" : " ");
  }
  return 0;
}
