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>
#pragma GCC optimize("O3")
using namespace std;
int add(int a, int b) {
  long long x = a + b;
  if (x >= 1000000007) x -= 1000000007;
  if (x < 0) x += 1000000007;
  return x;
}
long long mul(long long a, long long b) { return (a * b) % 1000000007; }
long long pw(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return ans;
}
long long n, k, v[2002], ans[2002], fact[4002], inv[4002], fact2[4002],
    inv2[4002];
long long comb(long long n, long long x) {
  long long ans = mul(fact2[n - (k - 1)], inv[x]);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  ++k;
  for (int i = 1; i <= n; ++i) cin >> v[i];
  fact[0] = inv[0] = 1;
  fact2[0] = inv2[0] = 1;
  for (int i = 1; i <= 4000; ++i) {
    fact[i] = mul(fact[i - 1], i);
    inv[i] = pw(fact[i], 1000000007 - 2);
    fact2[i] = mul(fact2[i - 1], k - 2 + i);
    inv2[i] = pw(fact2[i], 1000000007 - 2);
  }
  for (int i = 1; i <= n; ++i)
    for (int j = i; j <= n; ++j)
      ans[j] = add(ans[j], mul(v[i], comb(k - 1 + j - i, j - i)));
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  return 0;
}
