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 N = 2e3 + 5;
const int mod = 1e9 + 7;
inline int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
inline int sub(int x, int y) {
  x -= y;
  if (x < 0) x += mod;
  return x;
}
inline int mult(int x, int y) { return x * 1ll * y % mod; }
inline int fast(int b, int e) {
  if (!e) return 1;
  if (e & 1) return mult(b, fast(mult(b, b), e >> 1));
  return fast(mult(b, b), e >> 1);
}
inline int divi(int x, int y) { return x * 1ll * fast(y, mod - 2) % mod; }
int Fact[N], iFact[N];
void init() {
  Fact[0] = Fact[1] = iFact[0] = iFact[1] = 1;
  for (int i = 2; i < N; i++) {
    Fact[i] = mult(Fact[i - 1], i);
    iFact[i] = fast(Fact[i], mod - 2);
  }
}
unordered_map<int, unordered_map<int, int> > mp;
int C(int n, int k) { return mp[n][k]; }
int C2(int n, int k) {
  int up = 1;
  for (int i = k + 1; i <= n; i++) up = mult(up, i);
  return mult(up, iFact[n - k]);
}
int n, k;
int a[N];
int sv[N][N];
void go() {
  for (int i = 0; i <= n; i++) mp[k + i - 1][k - 1] = C2(k + i - 1, k - 1);
}
void brute(int n, int k) {
  for (int i = 0; i < n; i++) {
    sv[i][i] = 1;
  }
  for (int tmp = 0; tmp < k; tmp++) {
    for (int i = n - 1; i >= 0; i--) {
      for (int j = 0; j < i; j++) {
        for (int f = 0; f < n; f++) sv[i][f] += sv[j][f];
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << sv[i][j] << ' ';
    }
    puts("");
  }
}
int main() {
  init();
  scanf("%d%d", &n, &k);
  go();
  if (k <= n) {
    for (int i = 0; i < n; i++) scanf("%d", a + i);
    while (k--) {
      for (int i = 1; i < n; i++) a[i] = add(a[i], a[i - 1]);
    }
    for (int i = 0; i < n; i++) printf("%d%c", a[i], (i + 1 == n ? '\n' : ' '));
    return 0;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    int out = 0;
    int idx = k - 1;
    for (int j = i; j >= 0; j--, idx++) {
      out = add(out, mult(C(idx, k - 1), a[j]));
    }
    printf("%d%c", out, (i + 1 == n ? '\n' : ' '));
  }
  return 0;
}
