#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 20;
long long dp[maxn][10], pd[maxn][10], sz[maxn], m[maxn], k, res, x[maxn];
vector<int> adj[maxn];
int mod(int a) { return (a + k) % k; }
void dfs(int v, int p = -1) {
  sz[v]++;
  for (auto u : adj[v]) {
    if (u == p) continue;
    dfs(u, v);
    dp[v][1]++;
    for (int i = 0; i < k; i++) dp[v][i] += dp[u][mod(i - 1)];
    x[v] += x[u] + sz[u];
    sz[v] += sz[u];
  }
  for (auto u : adj[v]) {
    if (u == p) continue;
    res += (x[u] + sz[u]) * (sz[v] - sz[u]);
  }
}
void sfd(int v, int p = -1) {
  if (p != -1) {
    pd[v][1]++;
    for (int i = 0; i < k; i++) {
      pd[v][i] +=
          pd[p][mod(i - 1)] + dp[p][mod(i - 1)] - dp[v][mod(mod(i - 1) - 1)];
      if (i == 2 || (k == 2 && !i)) pd[v][i]--;
    }
  }
  for (auto u : adj[v])
    if (u != p) sfd(u, v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[--a].push_back(--b);
    adj[b].push_back(a);
  }
  dfs(0);
  sfd(0);
  long long sum = 0, z = 0;
  for (int i = 1; i < k; i++) {
    sum = 0;
    for (int j = 0; j < n; j++) {
      sum += dp[j][i] + pd[j][i];
    }
    sum /= 2;
    z += sum;
    res -= sum * i;
  }
  res /= k;
  res += z;
  cout << res;
}

### Prompt

Create a solution in cpp for 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 maxn = 2e5 + 20;
long long dp[maxn][10], pd[maxn][10], sz[maxn], m[maxn], k, res, x[maxn];
vector<int> adj[maxn];
int mod(int a) { return (a + k) % k; }
void dfs(int v, int p = -1) {
  sz[v]++;
  for (auto u : adj[v]) {
    if (u == p) continue;
    dfs(u, v);
    dp[v][1]++;
    for (int i = 0; i < k; i++) dp[v][i] += dp[u][mod(i - 1)];
    x[v] += x[u] + sz[u];
    sz[v] += sz[u];
  }
  for (auto u : adj[v]) {
    if (u == p) continue;
    res += (x[u] + sz[u]) * (sz[v] - sz[u]);
  }
}
void sfd(int v, int p = -1) {
  if (p != -1) {
    pd[v][1]++;
    for (int i = 0; i < k; i++) {
      pd[v][i] +=
          pd[p][mod(i - 1)] + dp[p][mod(i - 1)] - dp[v][mod(mod(i - 1) - 1)];
      if (i == 2 || (k == 2 && !i)) pd[v][i]--;
    }
  }
  for (auto u : adj[v])
    if (u != p) sfd(u, v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[--a].push_back(--b);
    adj[b].push_back(a);
  }
  dfs(0);
  sfd(0);
  long long sum = 0, z = 0;
  for (int i = 1; i < k; i++) {
    sum = 0;
    for (int j = 0; j < n; j++) {
      sum += dp[j][i] + pd[j][i];
    }
    sum /= 2;
    z += sum;
    res -= sum * i;
  }
  res /= k;
  res += z;
  cout << res;
}

```