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;
int n, k, pos;
long long tmp[2005], in[2005], dp[2005], mat[2005][2005], cnt[2005][2005],
    ans[2005];
long long pw(long long x, long long y) {
  long long re = 1;
  if (y == 0) return 1;
  re = pw(x, y / 2);
  re = (re * re) % 1000000007ll;
  if (y & 1) re = (re * x) % 1000000007ll;
  return re;
}
long long c(long long x, long long y) {
  if (dp[x - y] != -1) return dp[x - y];
  long long re = 1;
  pos = 0;
  for (long long i = x; i >= y + 1; i--) re = (re * i) % 1000000007ll;
  for (int i = 2; i <= x - y; i++)
    re = (re * pw(i, 1000000007ll - 2)) % 1000000007ll;
  return dp[x - y] = re;
}
int main() {
  memset(dp, -1, sizeof(dp));
  while (~scanf("%d %d", &n, &k)) {
    memset(ans, 0, sizeof(ans));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) scanf("%I64d", &in[i]);
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        if (j >= i)
          mat[i][j] = c(j - i + k - 1, k - 1);
        else
          mat[i][j] = 0;
      }
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) cnt[j][i] += mat[i][j];
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        ans[i] += in[j] * cnt[i][j];
        ans[i] %= 1000000007ll;
      }
    for (int i = 1; i <= n; i++) {
      if (i - 1) putchar(' ');
      printf("%I64d", ans[i]);
    }
    puts("");
  }
  return 0;
}
