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;
long long mod = 1000000007;
long long inv[3000];
long long inverse(long long a) {
  long long g = a;
  if (inv[a] != -1) return inv[a];
  long long b = mod - 2;
  long long c = 1;
  while (b > 0) {
    if (b & 1) c = (c * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  inv[g] = c;
  return inv[g];
}
long long degs[3000];
long long a[3000];
long long k;
int n;
long long nad(long long x, long long y) {
  long long r = 1;
  for (long long i = 1; i <= y; i++) {
    r = (((r * (x - i + 1)) % mod) * inverse(i)) % mod;
  }
  return r;
}
void calc() {
  if (k == 0) {
    degs[0] = 1;
    return;
  } else {
    degs[0] = 1;
    for (int i = 1; i < n; i++) {
      degs[i] = nad(k - 1 + i, i);
    }
  }
}
int main() {
  scanf("%d %lld", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    inv[i + 1] = -1;
  }
  calc();
  long long s;
  for (int i = 0; i < n; i++) {
    s = 0;
    for (int j = 0; j <= i; j++) s += (degs[j] * a[i - j]) % mod;
    s %= mod;
    printf("%lld ", s);
  }
  return 0;
}
