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 N = 2e3 + 5, Mod = 1e9 + 7, MAXN = 3e5 + 10, LG = 25,
                M = 1e7 + 1, P = 727, Sq = 400;
const long long inf = 3e18 + 10;
long long dp[N], ans[N], fac[N], inv[N], a[N], n, k;
long long pw(long long c, long long d) {
  if (!d) return 1;
  long long ret = pw(c, d / 2);
  if (d % 2) return 1ll * ret * ret % Mod * c % Mod;
  return 1ll * ret * ret % Mod;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  if (k < 2) {
    while (k) {
      for (long long i = 1; i <= n; i++) a[i] = a[i - 1] + a[i];
      k--;
    }
    for (long long i = 1; i <= n; i++) cout << a[i] << " ";
    return 0;
  }
  dp[1] = k;
  for (long long i = 2; i < n; i++)
    dp[i] = (1ll * dp[i - 1] * (k - 1 + i) % Mod);
  fac[0] = 1;
  for (long long i = 1; i < n; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
  inv[n - 1] = pw(fac[n - 1], Mod - 2);
  for (long long i = n - 2; i >= 0; i--)
    inv[i] = 1ll * inv[i + 1] * (i + 1) % Mod;
  for (long long i = 1; i <= n; i++)
    for (long long j = i + 1; j <= n; j++)
      ans[j] = (ans[j] + 1ll * dp[j - i] * inv[j - i] % Mod * a[i] % Mod) % Mod;
  for (long long i = 1; i <= n; i++) cout << (ans[i] + a[i]) % Mod << " ";
  return 0;
}
