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 = 1000000007;
long long C[2005][2005];
long long a[2005], s[2005];
void gcd(long long a, long long b, long long &d, long long &x, long long &y) {
  if (!b) {
    d = a, x = 1, y = 0;
  } else {
    gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inv(long long a, long long n) {
  long long d, x, y;
  gcd(a, n, d, x, y);
  return d == 1 ? (x + n) % n : -1;
}
void init(int n, int K) {
  C[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    long long t = inv(i, mod);
    assert(t != -1);
    t = t * (K - i + 1) % mod;
    C[0][i] = t * C[0][i - 1] % mod;
  }
  for (int i = 1; i < n; i++) {
    for (int j = i; j < n; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
}
int main() {
  long long n, k;
  scanf("%I64d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
  }
  if (k == 0) {
    for (int i = 1; i <= n; i++) {
      printf("%I64d ", a[i]);
    }
    return 0;
  }
  init(n, k - 1);
  for (int i = 1; i <= n; i++) {
    for (int j = i; j <= n; j++) {
      s[j] = (s[j] + C[j - i][j - i] * a[i] % mod) % mod;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%I64d ", s[i]);
  }
  printf("\n");
  return 0;
}
