#include <bits/stdc++.h>
using namespace std;
long long num[200005][6];
int n, k;
int siz[200005];
long long dp[200005];
vector<int> g[200005];
void dfs1(int u, int fa) {
  siz[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    for (int j = 1; j < k; j++) {
      num[u][j + 1] += num[v][j];
    }
    num[u][0] += num[v][0];
    num[u][0] += num[v][k];
    num[u][1] += num[v][k];
    num[u][1]++;
  }
}
void dfs2(int u, int fa) {
  if (fa == 0) {
    for (int i = 0; i <= k; i++) {
      dp[u] += num[u][i];
    }
  } else {
    long long a[10];
    for (int i = 0; i <= k; i++) a[i] = num[fa][i];
    for (int j = 1; j < k; j++) {
      a[j + 1] -= num[u][j];
    }
    a[0] -= num[u][0];
    a[0] -= num[u][k];
    a[1] -= num[u][k];
    a[1]--;
    for (int j = 1; j < k; j++) {
      num[u][j + 1] += a[j];
    }
    num[u][0] += a[0];
    num[u][0] += a[k];
    num[u][1] += a[k];
    num[u][1]++;
    for (int i = 0; i <= k; i++) {
      dp[u] += num[u][i];
    }
  }
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs2(v, u);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += dp[i];
  }
  cout << ans / 2 << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

A tree is an undirected connected graph without cycles. The distance between two vertices is the number of edges in a simple path between them.

Limak is a little polar bear. He lives in a tree that consists of n vertices, numbered 1 through n.

Limak recently learned how to jump. He can jump from a vertex to any vertex within distance at most k.

For a pair of vertices (s, t) we define f(s, t) as the minimum number of jumps Limak needs to get from s to t. Your task is to find the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.

Input

The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ 5) — the number of vertices in the tree and the maximum allowed jump distance respectively.

The next n - 1 lines describe edges in the tree. The i-th of those lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) — the indices on vertices connected with i-th edge.

It's guaranteed that the given edges form a tree.

Output

Print one integer, denoting the sum of f(s, t) over all pairs of vertices (s, t) such that s < t.

Examples

Input

6 2
1 2
1 3
2 4
2 5
4 6


Output

20


Input

13 3
1 2
3 2
4 2
5 2
3 6
10 6
6 7
6 13
5 8
5 9
9 11
11 12


Output

114


Input

3 5
2 1
3 1


Output

3

Note

In the first sample, the given tree has 6 vertices and it's displayed on the drawing below. Limak can jump to any vertex within distance at most 2. For example, from the vertex 5 he can jump to any of vertices: 1, 2 and 4 (well, he can also jump to the vertex 5 itself).

<image>

There are <image> pairs of vertices (s, t) such that s < t. For 5 of those pairs Limak would need two jumps: (1, 6), (3, 4), (3, 5), (3, 6), (5, 6). For other 10 pairs one jump is enough. So, the answer is 5·2 + 10·1 = 20.

In the third sample, Limak can jump between every two vertices directly. There are 3 pairs of vertices (s < t), so the answer is 3·1 = 3.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long num[200005][6];
int n, k;
int siz[200005];
long long dp[200005];
vector<int> g[200005];
void dfs1(int u, int fa) {
  siz[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    for (int j = 1; j < k; j++) {
      num[u][j + 1] += num[v][j];
    }
    num[u][0] += num[v][0];
    num[u][0] += num[v][k];
    num[u][1] += num[v][k];
    num[u][1]++;
  }
}
void dfs2(int u, int fa) {
  if (fa == 0) {
    for (int i = 0; i <= k; i++) {
      dp[u] += num[u][i];
    }
  } else {
    long long a[10];
    for (int i = 0; i <= k; i++) a[i] = num[fa][i];
    for (int j = 1; j < k; j++) {
      a[j + 1] -= num[u][j];
    }
    a[0] -= num[u][0];
    a[0] -= num[u][k];
    a[1] -= num[u][k];
    a[1]--;
    for (int j = 1; j < k; j++) {
      num[u][j + 1] += a[j];
    }
    num[u][0] += a[0];
    num[u][0] += a[k];
    num[u][1] += a[k];
    num[u][1]++;
    for (int i = 0; i <= k; i++) {
      dp[u] += num[u][i];
    }
  }
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == fa) continue;
    dfs2(v, u);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += dp[i];
  }
  cout << ans / 2 << endl;
  return 0;
}

```