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 MAXN = 2e3 + 10, md = 1e9 + 7;
struct matrix {
  long long a[MAXN];
  matrix(long long x = 1) {
    for (long long i = 0; i < MAXN; ++i) a[i] = x;
  }
};
matrix operator*(matrix& m1, matrix& m2) {
  matrix res(0);
  for (long long i = 0; i < MAXN; ++i)
    for (long long j = 0; j <= i; ++j) {
      res.a[i] += (m1.a[j] * m2.a[i - j]) % md;
      if (res.a[i] >= md) res.a[i] %= md;
    }
  return res;
}
matrix operator^(matrix& a, long long p) {
  matrix res(0);
  res.a[0] = 1;
  for (; p; p >>= 1, a = a * a)
    if (p & 1) res = res * a;
  return res;
}
long long a[MAXN];
int32_t main() {
  long long n, k;
  cin >> n >> k;
  matrix now;
  now = now ^ k;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    long long ans = 0;
    for (long long j = 0; j <= i; ++j) {
      ans += (a[j] * now.a[i - j]) % md;
      if (ans >= md) ans %= md;
    }
    cout << ans << ' ';
  }
  return 0;
}
