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;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long _INF = 0xc0c0c0c0c0c0c0c0;
const long long mod = (int)1e9 + 7;
const int N = 2e3 + 100;
int n, k;
int a[N], b[N];
int inv[N];
int c[N][N];
int init() {
  int MOD = mod;
  inv[1] = 1;
  for (int i = 2; i < N; i++) {
    inv[i] = (MOD - MOD / i) * 1ll * inv[MOD % i] % MOD;
  }
  c[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    c[i][0] = 1;
    int now = k + i - 1;
    for (int j = 1; j <= 2; ++j) {
      c[i][j] = 1ll * c[i][j - 1] * now % mod * inv[j] % mod;
      --now;
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  init();
  if (k) {
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= i; ++j) {
        b[i] = (b[i] + a[j] * c[i - j][i - j]) % mod;
      }
    }
    for (int i = 1; i <= n; ++i) a[i] = b[i];
  }
  for (int i = 1; i <= n; ++i) {
    printf("%d%c", a[i], " \n"[i == n]);
  }
  return 0;
}
