#include <bits/stdc++.h>
using namespace std;
long long ans = 0;
long long subtree[200005], treesize, saltos, qts[5], k;
bool vis[200005];
vector<int> G[200005];
void cnt(long long v, long long p) {
  subtree[v] = 1;
  for (long long x : G[v])
    if (x != p && !vis[x]) {
      cnt(x, v);
      subtree[v] += subtree[x];
    }
}
long long find_centroid(long long v, long long p) {
  for (long long x : G[v]) {
    if (x != p && !vis[x]) {
      if (subtree[x] * 2 > treesize) return find_centroid(x, v);
    }
  }
  return v;
}
void find_ans(long long v, long long p, long long d_atual) {
  long long meu_salto = d_atual / k;
  long long sobra = d_atual % k;
  ans += saltos;
  for (long long i = 0; i < k; i++) {
    if (i + sobra == 0)
      ans += (long long)qts[i] * meu_salto;
    else if (i + sobra <= k)
      ans += (long long)qts[i] * (meu_salto + 1);
    else
      ans += (long long)qts[i] * (meu_salto + 2);
  }
  for (long long x : G[v]) {
    if (x != p && !vis[x]) find_ans(x, v, d_atual + 1);
  }
}
void atualiza(long long v, long long p, long long d_atual) {
  long long meu_salto = d_atual / k;
  long long sobra = d_atual % k;
  saltos += meu_salto;
  qts[sobra]++;
  for (long long x : G[v]) {
    if (x != p && !vis[x]) atualiza(x, v, d_atual + 1);
  }
}
void decomp(long long v, long long p) {
  cnt(v, v);
  treesize = subtree[v];
  long long cent = find_centroid(v, v);
  saltos = 0;
  for (long long i = 0; i < k; i++) qts[i] = 0;
  qts[0] = 1;
  vis[cent] = 1;
  for (long long x : G[cent]) {
    if (!vis[x]) {
      find_ans(x, cent, 1);
      atualiza(x, cent, 1);
    }
  }
  for (long long x : G[cent]) {
    if (!vis[x]) decomp(x, cent);
  }
}
int main() {
  long long n;
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i < n; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  decomp(1, 1);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
long long ans = 0;
long long subtree[200005], treesize, saltos, qts[5], k;
bool vis[200005];
vector<int> G[200005];
void cnt(long long v, long long p) {
  subtree[v] = 1;
  for (long long x : G[v])
    if (x != p && !vis[x]) {
      cnt(x, v);
      subtree[v] += subtree[x];
    }
}
long long find_centroid(long long v, long long p) {
  for (long long x : G[v]) {
    if (x != p && !vis[x]) {
      if (subtree[x] * 2 > treesize) return find_centroid(x, v);
    }
  }
  return v;
}
void find_ans(long long v, long long p, long long d_atual) {
  long long meu_salto = d_atual / k;
  long long sobra = d_atual % k;
  ans += saltos;
  for (long long i = 0; i < k; i++) {
    if (i + sobra == 0)
      ans += (long long)qts[i] * meu_salto;
    else if (i + sobra <= k)
      ans += (long long)qts[i] * (meu_salto + 1);
    else
      ans += (long long)qts[i] * (meu_salto + 2);
  }
  for (long long x : G[v]) {
    if (x != p && !vis[x]) find_ans(x, v, d_atual + 1);
  }
}
void atualiza(long long v, long long p, long long d_atual) {
  long long meu_salto = d_atual / k;
  long long sobra = d_atual % k;
  saltos += meu_salto;
  qts[sobra]++;
  for (long long x : G[v]) {
    if (x != p && !vis[x]) atualiza(x, v, d_atual + 1);
  }
}
void decomp(long long v, long long p) {
  cnt(v, v);
  treesize = subtree[v];
  long long cent = find_centroid(v, v);
  saltos = 0;
  for (long long i = 0; i < k; i++) qts[i] = 0;
  qts[0] = 1;
  vis[cent] = 1;
  for (long long x : G[cent]) {
    if (!vis[x]) {
      find_ans(x, cent, 1);
      atualiza(x, cent, 1);
    }
  }
  for (long long x : G[cent]) {
    if (!vis[x]) decomp(x, cent);
  }
}
int main() {
  long long n;
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i < n; i++) {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  decomp(1, 1);
  printf("%lld\n", ans);
  return 0;
}

```