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>
const long long mod = 1e9 + 7;
using namespace std;
const long long mod2 = 998244353;
int gcdExtended(int a, int b, int* x, int* y);
int modInverse(int a, int m) {
  int x, y;
  int g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return -1;
  else {
    int res = (x % m + m) % m;
    return res;
  }
}
int gcdExtended(int a, int b, int* x, int* y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  vector<long long> coeff(n);
  coeff[0] = 1;
  long long prev_coeff = 1;
  for (long long i = 1; i < n; i++) {
    long long mi = modInverse(i, mod);
    long long temp =
        ((((prev_coeff % mod) * ((k + i - 1) % mod)) % mod) * (mi % mod)) % mod;
    coeff[i] = temp;
    prev_coeff = temp;
    if (temp >= mod) cout << "hi";
  }
  for (long long i = 0; i < n; i++) {
    long long ans = 0;
    for (long long j = 0; j <= i; j++) {
      ans = ((ans) + (((a[j] % mod) * (coeff[i - j] % mod)) % mod)) % mod;
    }
    if (ans >= mod) cout << "hians";
    cout << ans << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
