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 int SIZE = 1E6 + 5;
const int INF = 0x3f3f3f3f;
const long long INFll = 0x3f3f3f3f3f3f3f3f;
const long long fat = 1E9 + 7;
int n;
vector<long long> matMul(vector<long long> a, vector<long long> b) {
  vector<long long> m(n);
  for (int i = 0; i < n; i++) {
    m[i] = 0;
    for (int j = 0; j <= i; j++) {
      m[i] += a[i - j] * b[j];
      m[i] %= fat;
    }
  }
  return m;
};
vector<long long> matPow(vector<long long> resposta, int p) {
  vector<long long> base(n);
  for (int i = 0; i < n; i++) {
    base[i] = 1;
  }
  while (p) {
    if (p & 1) resposta = matMul(resposta, base);
    base = matMul(base, base);
    p >>= 1;
  }
  return resposta;
}
int main() {
  int k;
  scanf("%d %d", &n, &k);
  vector<long long> resposta(n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    resposta[i] = x;
  }
  resposta = matPow(resposta, k);
  for (int i = 0; i < n; i++) {
    printf("%lld ", resposta[i]);
  }
  return 0;
}
