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 = 200100, MOD = 1e9 + 7;
int n, k;
long long A[MAXN], L[MAXN], ans[MAXN], INV[MAXN], DOS[MAXN];
vector<long long> P;
long long pot(long long x, long long e) {
  P.clear();
  long long p = 0;
  P.push_back(x);
  while (2 * DOS[p] <= e) {
    x = (x * x) % MOD;
    P.push_back(x);
    p++;
  }
  long long ans = 1;
  while (e > 0) {
    while (DOS[p] > e) p--;
    e -= DOS[p];
    ans = (ans * P[p]) % MOD;
  }
  return ans;
}
int main() {
  DOS[0] = 1;
  for (int i = 1; 2 * DOS[i - 1] < MOD; i++) {
    DOS[i] = 2 * DOS[i - 1];
  }
  for (int i = (int)(1); i < (int)(3000); i++) INV[i] = pot(i, MOD - 2) % MOD;
  while (cin >> n >> k) {
    for (int i = 0; i < (int)(n); i++) scanf("%d", &A[i]);
    L[0] = 1;
    for (int i = (int)(1); i < (int)(n); i++)
      L[i] = (L[i - 1] * (k + i - 1) % MOD) * INV[i] % MOD;
    for (int i = 0; i < (int)(n); i++) {
      long long sum = 0;
      for (int j = 0; j < (int)(i + 1); j++)
        sum = (sum + (A[j] * L[i - j] % MOD)) % MOD;
      ans[i] = sum;
    }
    for (int i = 0; i < (int)(n); i++) {
      if (i) printf(" ");
      printf("%lld", ans[i]);
    }
    puts("");
  }
  return 0;
}
