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 maxn = 2000 + 10, mod = 1e9 + 7;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int sub(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
inline int mul(int a, int b) { return a * 1ll * b % mod; }
int fpow(int a, int n) {
  int ret = 1;
  for (; n; n >>= 1) {
    if (n & 1) ret = mul(ret, a);
    a = mul(a, a);
  }
  return ret;
}
int a[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  vector<int> coef;
  coef.push_back(1);
  for (int i = 1; i <= n; ++i) {
    int s = 0;
    size_t ptr = 0;
    for (int j = i; j >= 1; --j) s = add(s, mul(a[j], coef[ptr++]));
    cout << s << ' ';
    coef.push_back(mul(coef.back(), mul(k + i - 1, fpow(i, mod - 2))));
  }
  return 0;
}
