#include <bits/stdc++.h>
using namespace std;
const int inf = 200009;
vector<int> adj[inf];
long long dp[5];
int num[5];
int com;
int k, n;
long long sol = 0;
int sz[inf];
bool del[inf];
int selsz;
int calsize(int x, int p) {
  sz[x] = 1;
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) sz[x] += calsize(adj[x][i], x);
  return sz[x];
}
int find_mid(int x, int p) {
  int mx = selsz - sz[x];
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) mx = max(mx, sz[adj[x][i]]);
  if (mx <= selsz / 2) return x;
  mx = -1;
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) mx = max(mx, find_mid(adj[x][i], x));
  return mx;
}
void calc(int x, int p, int d) {
  if (com == 1) {
    sol += (d + k - 1) / k;
    int sel = d / k, md = d % k;
    for (int i = 0; i < k; ++i) {
      sol += 1ll * sel * num[i];
      sol += dp[i];
      sol += 1ll * (md + i + k - 1) / k * num[i];
    }
  } else if (com == 2)
    dp[d % k] += d / k, num[d % k]++;
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) calc(adj[x][i], x, d + 1);
}
void solve(int x) {
  calsize(x, -1);
  selsz = sz[x];
  x = find_mid(x, -1);
  for (int i = 0; i < adj[x].size(); ++i) {
    if (del[adj[x][i]]) continue;
    com = 1;
    calc(adj[x][i], x, 1);
    com = 2;
    calc(adj[x][i], x, 1);
  }
  for (int i = 0; i < k; ++i) dp[i] = num[i] = 0;
  del[x] = true;
  for (int i = 0; i < adj[x].size(); ++i)
    if (!del[adj[x][i]]) solve(adj[x][i]);
}
int main() {
  scanf("%d%d", &n, &k);
  int a, b;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  solve(1);
  printf("%lld\n", sol);
}

### Prompt

Your challenge is to write a CPP solution 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 inf = 200009;
vector<int> adj[inf];
long long dp[5];
int num[5];
int com;
int k, n;
long long sol = 0;
int sz[inf];
bool del[inf];
int selsz;
int calsize(int x, int p) {
  sz[x] = 1;
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) sz[x] += calsize(adj[x][i], x);
  return sz[x];
}
int find_mid(int x, int p) {
  int mx = selsz - sz[x];
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) mx = max(mx, sz[adj[x][i]]);
  if (mx <= selsz / 2) return x;
  mx = -1;
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) mx = max(mx, find_mid(adj[x][i], x));
  return mx;
}
void calc(int x, int p, int d) {
  if (com == 1) {
    sol += (d + k - 1) / k;
    int sel = d / k, md = d % k;
    for (int i = 0; i < k; ++i) {
      sol += 1ll * sel * num[i];
      sol += dp[i];
      sol += 1ll * (md + i + k - 1) / k * num[i];
    }
  } else if (com == 2)
    dp[d % k] += d / k, num[d % k]++;
  for (int i = 0; i < adj[x].size(); ++i)
    if (adj[x][i] != p && !del[adj[x][i]]) calc(adj[x][i], x, d + 1);
}
void solve(int x) {
  calsize(x, -1);
  selsz = sz[x];
  x = find_mid(x, -1);
  for (int i = 0; i < adj[x].size(); ++i) {
    if (del[adj[x][i]]) continue;
    com = 1;
    calc(adj[x][i], x, 1);
    com = 2;
    calc(adj[x][i], x, 1);
  }
  for (int i = 0; i < k; ++i) dp[i] = num[i] = 0;
  del[x] = true;
  for (int i = 0; i < adj[x].size(); ++i)
    if (!del[adj[x][i]]) solve(adj[x][i]);
}
int main() {
  scanf("%d%d", &n, &k);
  int a, b;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  solve(1);
  printf("%lld\n", sol);
}

```