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 maxn = 2100;
const long long mod = 1e9 + 7;
long long c[maxn][maxn];
long long a[maxn];
long long n, k;
long long pow(long long a, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ans;
}
long long niYuan(long long a) { return pow(a, mod - 2); }
int main() {
  while (cin >> n >> k) {
    for (int i = 1; i <= n; ++i) cin >> a[i];
    c[1][0] = 1;
    for (int i = 2; i <= 2000; ++i) {
      long long tmpa = 1, tmpb = 1;
      for (int j = 1; j < i; ++j) {
        (tmpa *= (k + i - 3 - (j - 1))) %= mod;
        (tmpb *= j) %= mod;
      }
      c[i][0] = tmpa * niYuan(tmpb) % mod;
    }
    c[1][1] = 1LL;
    for (int i = 2; i <= n; ++i) {
      for (int j = i; j > 1; --j) c[i][j] = c[i - 1][j - 1];
      c[i][1] = (c[i][0] + c[i - 1][1]) % mod;
    }
    for (int i = 1; i <= n; ++i) {
      long long ans = 0;
      for (int j = 1; j <= i; ++j) ans = (ans + c[i][j] * a[j] % mod) % mod;
      cout << ans << " ";
    }
    cout << endl;
  }
  return 0;
}
