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;
int n, k;
long long B[2000][2000];
int a[2000];
long long mod = 1000 * 1000 * 1000 + 7;
long long invFact[2001];
long long calcC(int n, int k) {
  long long ans = 1;
  for (int i = n - k + 1; i <= n; i++) ans = (ans * i) % mod;
  return (ans * invFact[k]) % mod;
}
long long binpow(long long x, int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return x;
  else if (n & 1)
    return (binpow(x, n - 1) * x) % mod;
  else
    return binpow((x * x) % mod, n >> 1);
}
long long inv(long long x) { return binpow(x, mod - 2); }
void pre() {
  invFact[0] = 1;
  invFact[1] = 1;
  for (int i = 2; i <= 2000; i++) {
    invFact[i] = invFact[i - 1] * inv(i);
    invFact[i] %= mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  pre();
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (k == 0) {
    for (int i = 0; i < n; i++) {
      cout << a[i] % mod << " ";
    }
    return 0;
  }
  if (k == 1) {
    int curr = 0;
    for (int i = 0; i < n; i++) {
      curr = (curr + a[i]) % mod;
      cout << curr << " ";
    }
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      for (int j = 0; j < n - i; j++) {
        B[i + j][i] = calcC(k + j - 1, j);
      }
    } else {
      for (int j = 0; j < n - i; j++) {
        B[i + j][i] = B[j][0];
      }
    }
  }
  for (int i = 0; i < n; i++) {
    long long ans = 0;
    for (int j = 0; j < n; j++) {
      ans = (ans + B[i][j] * a[j]) % mod;
    }
    cout << ans << " ";
  }
  return 0;
}
