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;
int totalCases, testNum;
const long long MOD = 1000000007;
int N, K;
vector<long long> mult(vector<long long> A, vector<long long> B) {
  vector<long long> ret;
  int n = A.size();
  ret.resize(n);
  for (int i = 0; i < n; i++) {
    ret[i] = 0;
    for (int k = 0; k <= i; k++) {
      long long term = 0;
      term = (A[i - k] * B[k]) % MOD;
      ret[i] = (ret[i] + term) % MOD;
    }
  }
  return ret;
}
vector<long long> exp(vector<long long> cur, int e) {
  if (e == 1) return cur;
  vector<long long> ret, rec;
  rec = exp(mult(cur, cur), e >> 1);
  if (e & 1)
    ret = mult(rec, cur);
  else
    ret = rec;
  return ret;
}
vector<long long> a;
void preprocess() {}
bool input() {
  scanf("%d", &N);
  scanf("%d", &K);
  long long x;
  for (int i = 0; i < N; i++) {
    cin >> x;
    a.push_back(x);
  }
  return true;
}
void solve() {
  if (K == 0) {
    for (int i = 0; i < N; i++) cout << a[i] << " ";
    cout << endl;
    return;
  }
  vector<long long> T;
  for (int i = 0; i < N; i++) T.push_back(1);
  T = exp(T, K);
  vector<long long> ret;
  ret.resize(N);
  for (int i = 0; i < N; i++) {
    ret[i] = 0;
    for (int j = 0; j <= i; j++)
      ret[i] = (ret[i] + ((T[i - j] * a[j]) % MOD)) % MOD;
  }
  for (int i = 0; i < N; i++) {
    cout << ret[i] << " ";
  }
  cout << endl;
}
int main() {
  preprocess();
  totalCases = 1;
  for (testNum = 1; testNum <= totalCases; testNum++) {
    if (!input()) break;
    solve();
  }
}
