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 comment(linker, "/stack:100000000")
using namespace std;
const int MOD = 1000000007;
long long mypow(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else if (b % 2 == 0) {
    long long t = mypow(a, b / 2);
    return t * t % MOD;
  } else {
    long long t = mypow(a, b - 1);
    return a * t % MOD;
  }
}
long long inv(long long x) { return mypow(x, MOD - 2); }
const int N = 2012;
long long a[N], b[N], c[N], f[N];
int main() {
  f[0] = 1;
  for (int i = 1; i < N; i++) {
    f[i] = f[i - 1] * i;
    f[i] %= MOD;
  }
  int n;
  long long k;
  cin >> n >> k;
  for (int(i) = 0; (i) < (n); (i)++) cin >> a[i];
  if (k == 0) {
    for (int(i) = 0; (i) < (n); (i)++) cout << a[i] << " ";
    return 0;
  }
  --k;
  b[0] = 1;
  for (int i = 1; i < N; i++) {
    b[i] = b[i - 1] * (k + i);
    b[i] %= MOD;
  }
  for (int(i) = 0; (i) < (N); (i)++) c[i] = b[i] * inv(f[i]) % MOD;
  for (int(i) = 0; (i) < (n); (i)++) {
    long long res = 0;
    for (int(j) = 0; (j) < (i + 1); (j)++) {
      res += c[i - j] * a[j];
      res %= MOD;
    }
    cout << res << " ";
  }
  return 0;
}
