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 c[2000 + 5][2000 + 5];
void GC() {
  memset(c, 0, sizeof(c));
  c[0][0] = 1;
  for (int i = 1; i <= 2000; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % 1000000007ll;
    }
  }
}
long long pw(long long x, long long t) {
  long long ret = 1 % 1000000007ll;
  while (t) {
    if (t & 1) ret = ret * x % 1000000007ll;
    x = x * x % 1000000007ll;
    t >>= 1;
  }
  return ret;
}
int n, m;
long long a[2000 + 5];
long long x[2000 + 5];
long long inv[2000 + 5];
int main() {
  for (int i = 0; i <= 2000; i++) inv[i] = pw(i, 1000000007ll - 2);
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
    if (m == 0) {
      for (int i = 1; i <= n; i++) {
        printf("%I64d%c", a[i], i == n ? '\n' : ' ');
      }
      continue;
    }
    long long u = 1, d = 1;
    x[0] = 1;
    for (int i = 1; i <= n; i++) {
      u = u * (m + i - 1) % 1000000007ll;
      d = d * inv[i] % 1000000007ll;
      x[i] = u * d % 1000000007ll;
    }
    for (int i = 1; i <= n; i++) {
      long long ans = 0;
      for (int j = 1; j <= i; j++) {
        ans += x[i - j] * a[j] % 1000000007ll;
        ans %= 1000000007ll;
      }
      printf("%I64d%c", ans, i == n ? '\n' : ' ');
    }
  }
  return 0;
}
