#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
long long f[maxn][10], g[maxn][10], n, res, k, rres[maxn], rvres[maxn],
    f1[maxn];
vector<int> adj[maxn];
void dfs(int u, int p) {
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == p) continue;
    dfs(v, u);
    for (int j = 0; j < k; ++j) f[u][j + 1] += f[v][j];
    rres[u] += rres[v] + 1;
    rres[u] += f[v][k];
    f[u][1] += f[v][k];
  }
  f[u][0] = 1;
}
void dfs2(int u, int p) {
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == p) continue;
    for (int j = 0; j < k; ++j) {
      g[v][j + 1] += g[u][j] + f[u][j];
      if (j > 0) g[v][j + 1] -= f[v][j - 1];
      if (j == 1) g[v][j + 1] -= f[v][k];
    }
    rvres[v] += rvres[u] + rres[u] - rres[v] - 1 - f[v][k] + 1;
    rvres[v] += g[u][k] + f[u][k] - f[v][k - 1];
    g[v][1] += g[u][k] + f[u][k] - f[v][k - 1];
    dfs2(v, u);
  }
}
void dfs1(int u, int p) {
  f1[u] = 1;
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == p) continue;
    dfs1(v, u);
    f1[u] += f1[v];
  }
  res += f1[u] * (n - f1[u]);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 1) {
    dfs1(1, 0);
    cout << res;
    return 0;
  }
  dfs(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= n; ++i) res += rres[i] + rvres[i];
  cout << res / 2;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int maxn = 2e5 + 7;
long long f[maxn][10], g[maxn][10], n, res, k, rres[maxn], rvres[maxn],
    f1[maxn];
vector<int> adj[maxn];
void dfs(int u, int p) {
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == p) continue;
    dfs(v, u);
    for (int j = 0; j < k; ++j) f[u][j + 1] += f[v][j];
    rres[u] += rres[v] + 1;
    rres[u] += f[v][k];
    f[u][1] += f[v][k];
  }
  f[u][0] = 1;
}
void dfs2(int u, int p) {
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == p) continue;
    for (int j = 0; j < k; ++j) {
      g[v][j + 1] += g[u][j] + f[u][j];
      if (j > 0) g[v][j + 1] -= f[v][j - 1];
      if (j == 1) g[v][j + 1] -= f[v][k];
    }
    rvres[v] += rvres[u] + rres[u] - rres[v] - 1 - f[v][k] + 1;
    rvres[v] += g[u][k] + f[u][k] - f[v][k - 1];
    g[v][1] += g[u][k] + f[u][k] - f[v][k - 1];
    dfs2(v, u);
  }
}
void dfs1(int u, int p) {
  f1[u] = 1;
  for (int i = 0; i < adj[u].size(); ++i) {
    int v = adj[u][i];
    if (v == p) continue;
    dfs1(v, u);
    f1[u] += f1[v];
  }
  res += f1[u] * (n - f1[u]);
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 1) {
    dfs1(1, 0);
    cout << res;
    return 0;
  }
  dfs(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= n; ++i) res += rres[i] + rvres[i];
  cout << res / 2;
}

```