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 int maxn = 2e3 + 10;
const int MOD = 1e9 + 7;
long long bit[maxn], n, k;
long long sum(int i) {
  long long s = 0;
  while (i > 0) {
    s += bit[i];
    s %= MOD;
    i -= i & -i;
  }
  return s;
}
void add(int i, long long x) {
  while (i <= n) {
    bit[i] += x;
    bit[i] %= MOD;
    i += i & -i;
  }
}
long long dp[maxn];
long long a[maxn];
long long quick_mod(long long a, long long b) {
  long long ans = 1;
  a %= MOD;
  while (b) {
    if (b & 1) {
      ans = ans * a % MOD;
      b--;
    }
    b >>= 1;
    a = a * a % MOD;
  }
  return ans;
}
long long C(long long n, long long m) {
  if (m > n) return 0;
  long long ans = 1;
  for (int i = 1; i <= m; i++) {
    long long a = (n + i - m) % MOD;
    long long b = i % MOD;
    ans = ans * (a * quick_mod(b, MOD - 2) % MOD) % MOD;
  }
  return ans;
}
long long Lucas(long long n, long long m) {
  if (m == 0) return 1;
  return C(n % MOD, m % MOD) * Lucas(n / MOD, m / MOD) % MOD;
}
long long combine[maxn];
void init_combine() {
  for (int i = 1; i <= n; i++) combine[i] = Lucas(k + i - 1, i);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  dp[1] = a[1];
  init_combine();
  for (int i = 2; i <= n; i++) {
    dp[i] = a[i];
    for (int j = 1; j <= i - 1; j++) {
      dp[i] += (combine[j] * a[i - j]) % MOD;
      dp[i] %= MOD;
    }
  }
  for (int i = 1; i <= n; i++) cout << dp[i] << ' ';
  cout << endl;
  return 0;
}
