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;
long long fac[2028];
long long invfac[2028];
long long expo(long long base, long long exponent, long long mod) {
  long long ans = 1;
  while (exponent != 0) {
    if ((exponent & 1) == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base = base * base;
    base %= mod;
    exponent >>= 1;
  }
  return ans % mod;
}
long long add(long long a, long long b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
long long sub(long long a, long long b) {
  return ((a % 1000000007) - (b % 1000000007) + 1000000007) % 1000000007;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fac[0] = 1;
  invfac[0] = 1;
  for (long long int i = 1; i < 2028; i++) {
    fac[i] = mul(fac[i - 1], i);
  }
  for (long long int i = 1; i < 2028; i++) {
    invfac[i] = expo(i, 1000000007 - 2, 1000000007);
  }
  long long n, k;
  cin >> n >> k;
  std::vector<long long> a(n);
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  std::vector<long long> res(n + 1);
  res[0] = 1;
  for (long long int i = 1; i < n + 1; i++) {
    res[i] = mul(res[i - 1], mul(k + i - 1, invfac[i]));
  }
  std::vector<long long> fin(n);
  for (long long int i = 0; i < n; i++) {
    long long cres = 0;
    long long curr = 0;
    for (long long j = i; j >= 0; j--) {
      cres = add(cres, mul(res[j], a[curr]));
      curr++;
    }
    fin[i] = cres;
  }
  for (auto i : fin) {
    cout << i << " ";
  }
}
