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 = 1e9 + 10;
const long long int INFLL = 1e18 + 10;
const long double EPS = 1e-8;
const long double EPSLD = 1e-14;
const long long int MOD = 1e9 + 7;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
struct Mod {
  unsigned n;
  Mod() {}
  Mod(long long int x) {
    if (x < 0)
      n = x % MOD + MOD;
    else
      n = x % MOD;
  }
};
Mod operator+(Mod a, Mod b) { return Mod(a.n + b.n); }
Mod operator+=(Mod &a, Mod b) { return a = a + b; }
Mod operator-(Mod a, Mod b) { return Mod(a.n + MOD - b.n); }
Mod operator-=(Mod &a, Mod b) { return a = a - b; }
Mod operator*(Mod a, Mod b) { return Mod((long long int)a.n * b.n); }
Mod operator*=(Mod &a, Mod b) { return a = a * b; }
Mod modpow(Mod x, long long int k) {
  Mod res = 1;
  while (k) {
    if (k & 1) res *= x;
    k /= 2;
    x *= x;
  }
  return res;
}
Mod inv(Mod a) { return modpow(a, MOD - 2); }
Mod operator/(Mod a, Mod b) { return Mod((long long int)a.n * inv(b).n); }
Mod operator/=(Mod &a, Mod b) { return a = a / b; }
const int MAX_N = 2010;
int n, k;
int aa[MAX_N];
Mod a[MAX_N];
Mod iv[MAX_N];
Mod cC(int a, int b) {
  Mod res = 1;
  for (int i = (0); i < (int)(b); i++) res *= a - i;
  for (int i = (0); i < (int)(b); i++) res *= iv[i + 1];
  return res;
}
Mod c[MAX_N];
int main() {
  for (int i = (0); i < (int)(MAX_N); i++) iv[i] = inv(i);
  scanf("%d %d", &n, &k);
  for (int i = (0); i < (int)(n); i++) scanf("%d", &aa[i]);
  for (int i = (0); i < (int)(n); i++) a[i] = aa[i];
  for (int i = (0); i < (int)(MAX_N); i++) c[i] = cC(k + i - 1, i);
  for (int i = (0); i < (int)(n); i++) {
    Mod res = 0;
    for (int j = (0); j < (int)(i + 1); j++) res += a[j] * c[i - j];
    printf("%d%c", res.n, i == n - 1 ? '\n' : ' ');
  }
  return 0;
}
