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;
void io() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int MAXN = 2e3 + 5;
const long long INF = 0x3f3f3f3f3f3f3f;
const long long prim = 1e9 + 7;
int n, k;
long long a[MAXN];
long long t[MAXN];
long long ans[MAXN];
long long factk[MAXN];
long long fact[MAXN], invf[MAXN];
long long ppow(long long base, long long t) {
  long long ans = 1;
  while (t) {
    if (t & 1) {
      ans = (ans * base) % prim;
    }
    base = (base * base) % prim;
    t >>= 1;
  }
  return ans;
}
void init() {
  fact[0] = 1;
  factk[0] = 1;
  for (int i = 1; i < n; i++) {
    fact[i] = (fact[i - 1] * i) % prim;
    factk[i] = (factk[i - 1] * (k + i - 1)) % prim;
  }
  invf[n - 1] = ppow(fact[n - 1], prim - 2);
  for (int i = n - 2; i >= 0; i--) {
    invf[i] = (invf[i + 1] * (i + 1)) % prim;
  }
  for (int i = 0; i < n; i++) {
    t[i] = (factk[i] * invf[i]) % prim;
  }
}
int main() {
  io();
  cin >> n >> k;
  init();
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int p = 0; p < n; p++) {
    for (int i = 0; i <= p; i++) {
      ans[p] = (ans[p] + a[i] * t[p - i]) % prim;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << ' ';
  }
}
