#include <bits/stdc++.h>
using namespace std;
mt19937 Rand(time(0));
const int maxn = 2e5 + 5;
const long long maxc = 1e9 + 7;
int dpInChild[maxn][6], dpOutChild[maxn][6];
int n, k;
long long distInChild[maxn], distOutChild[maxn];
vector<int> adj[maxn];
int child[maxn];
void dfsBottomUp(int u, int par = -1) {
  child[u] = 1;
  dpInChild[u][0]++;
  for (auto v : adj[u]) {
    if (v == par) continue;
    dfsBottomUp(v, u);
    child[u] += child[v];
    for (int i = (0); i <= (k - 1); ++i)
      dpInChild[u][i] += dpInChild[v][(i - 1 + k) % k];
    distInChild[u] += distInChild[v] + child[v];
  }
}
long long ans;
void dfsTopDown(int u, int par = -1) {
  long long res = distInChild[u] + distOutChild[u];
  for (int i = (1); i <= (k - 1); ++i) {
    ans += dpInChild[u][i] + dpOutChild[u][i];
    res -= (dpInChild[u][i] + dpOutChild[u][i]) * i;
  }
  ans += res / k;
  for (auto v : adj[u]) {
    if (v == par) continue;
    for (int i = (0); i <= (k - 1); ++i) {
      dpOutChild[v][i] += dpOutChild[u][(i - 1 + k) % k];
      dpOutChild[v][i] +=
          dpInChild[u][(i - 1 + k) % k] - dpInChild[v][(i - 2 + k * k) % k];
    }
    distOutChild[v] += distOutChild[u] + n - child[v] + distInChild[u] -
                       distInChild[v] - child[v];
    dfsTopDown(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfsBottomUp(1);
  dfsTopDown(1);
  cout << ans / 2;
  return 0;
}

### 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;
mt19937 Rand(time(0));
const int maxn = 2e5 + 5;
const long long maxc = 1e9 + 7;
int dpInChild[maxn][6], dpOutChild[maxn][6];
int n, k;
long long distInChild[maxn], distOutChild[maxn];
vector<int> adj[maxn];
int child[maxn];
void dfsBottomUp(int u, int par = -1) {
  child[u] = 1;
  dpInChild[u][0]++;
  for (auto v : adj[u]) {
    if (v == par) continue;
    dfsBottomUp(v, u);
    child[u] += child[v];
    for (int i = (0); i <= (k - 1); ++i)
      dpInChild[u][i] += dpInChild[v][(i - 1 + k) % k];
    distInChild[u] += distInChild[v] + child[v];
  }
}
long long ans;
void dfsTopDown(int u, int par = -1) {
  long long res = distInChild[u] + distOutChild[u];
  for (int i = (1); i <= (k - 1); ++i) {
    ans += dpInChild[u][i] + dpOutChild[u][i];
    res -= (dpInChild[u][i] + dpOutChild[u][i]) * i;
  }
  ans += res / k;
  for (auto v : adj[u]) {
    if (v == par) continue;
    for (int i = (0); i <= (k - 1); ++i) {
      dpOutChild[v][i] += dpOutChild[u][(i - 1 + k) % k];
      dpOutChild[v][i] +=
          dpInChild[u][(i - 1 + k) % k] - dpInChild[v][(i - 2 + k * k) % k];
    }
    distOutChild[v] += distOutChild[u] + n - child[v] + distInChild[u] -
                       distInChild[v] - child[v];
    dfsTopDown(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = (1); i <= (n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfsBottomUp(1);
  dfsTopDown(1);
  cout << ans / 2;
  return 0;
}

```