#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 87;
int n, k;
long long sz[N][5], sl[N][5], dp[N][5], up[N][5];
vector<int> g[N];
void dfs(int d, int p, int u) {
  sz[u][0] = 1;
  for (int v : g[u])
    if (v != p) {
      dfs(d + 1, u, v);
      for (int i = 0; i < k; ++i) {
        sl[u][(i + 1) % k] += sl[v][i] + sz[v][i];
        sz[u][(i + 1) % k] += sz[v][i];
      }
    }
}
void gdp(int p, int u) {
  if (p != -1) {
    for (int i = 0; i < k; ++i) {
      dp[u][i] = dp[p][(i - 1 + k) % k] + up[p][(i - 1 + k) % k] +
                 sl[p][(i - 1 + k) % k] - sl[u][(i - 2 + k + k) % k] -
                 sz[u][(i - 2 + k + k) % k] + sz[p][(i - 1 + k) % k] -
                 sz[u][(i - 2 + k + k) % k];
      up[u][i] = up[p][(i - 1 + k) % k] + sz[p][(i - 1 + k) % k] -
                 sz[u][(i - 2 + k + k) % k];
    }
  }
  for (int v : g[u])
    if (v != p) {
      gdp(u, v);
    }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1, 0);
  gdp(-1, 0);
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    --sz[i][0];
    for (int j = 0; j < k; ++j) {
      ans += (dp[i][j] - j * up[i][j] + sl[i][j] - j * sz[i][j]) / k +
             (j ? (up[i][j] + sz[i][j]) : 0);
    }
  }
  printf("%lld\n", ans / 2);
}

### Prompt

Please create a solution in CPP to the following problem:

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 N = 2e5 + 87;
int n, k;
long long sz[N][5], sl[N][5], dp[N][5], up[N][5];
vector<int> g[N];
void dfs(int d, int p, int u) {
  sz[u][0] = 1;
  for (int v : g[u])
    if (v != p) {
      dfs(d + 1, u, v);
      for (int i = 0; i < k; ++i) {
        sl[u][(i + 1) % k] += sl[v][i] + sz[v][i];
        sz[u][(i + 1) % k] += sz[v][i];
      }
    }
}
void gdp(int p, int u) {
  if (p != -1) {
    for (int i = 0; i < k; ++i) {
      dp[u][i] = dp[p][(i - 1 + k) % k] + up[p][(i - 1 + k) % k] +
                 sl[p][(i - 1 + k) % k] - sl[u][(i - 2 + k + k) % k] -
                 sz[u][(i - 2 + k + k) % k] + sz[p][(i - 1 + k) % k] -
                 sz[u][(i - 2 + k + k) % k];
      up[u][i] = up[p][(i - 1 + k) % k] + sz[p][(i - 1 + k) % k] -
                 sz[u][(i - 2 + k + k) % k];
    }
  }
  for (int v : g[u])
    if (v != p) {
      gdp(u, v);
    }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(0, -1, 0);
  gdp(-1, 0);
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    --sz[i][0];
    for (int j = 0; j < k; ++j) {
      ans += (dp[i][j] - j * up[i][j] + sl[i][j] - j * sz[i][j]) / k +
             (j ? (up[i][j] + sz[i][j]) : 0);
    }
  }
  printf("%lld\n", ans / 2);
}

```