#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200000 + 5;
vector<int> tree[MAXN];
int siz[MAXN], g[MAXN][5];
int k;
void dfs1(int u, int pa) {
  siz[u] = 1;
  g[u][0] = 1;
  for (auto v : tree[u]) {
    if (v == pa) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    for (int j = 0; j < k; ++j) g[u][(j + 1) % k] += g[v][j];
  }
}
void dfs2(int u, int pa) {
  for (auto v : tree[u]) {
    if (v == pa) continue;
    int tmp[5] = {0};
    for (int i = 0; i < k; ++i)
      tmp[(i + 1) % k] = g[u][i] - g[v][(i - 1 + k) % k];
    for (int i = 0; i < k; ++i) g[v][i] += tmp[i];
    dfs2(v, u);
  }
}
long long solve(int n) {
  dfs1(1, 1);
  dfs2(1, 1);
  long long s[5] = {0};
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < k; ++j) s[j] += g[i][j];
  for (int i = 1; i < k; ++i) s[i] /= 2;
  long long dis = 0, sum = 0;
  for (int i = 1; i <= n; ++i) dis += 1LL * siz[i] * (n - siz[i]);
  for (int i = 1; i < k; ++i) {
    dis -= 1LL * i * s[i];
    sum += s[i];
  }
  dis = dis / k + sum;
  return dis;
}
int main() {
  int n;
  scanf("%d %d", &n, &k);
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d %d", &u, &v);
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  printf("%lld\n", solve(n));
  return 0;
}

### Prompt

Your task is to create 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 MAXN = 200000 + 5;
vector<int> tree[MAXN];
int siz[MAXN], g[MAXN][5];
int k;
void dfs1(int u, int pa) {
  siz[u] = 1;
  g[u][0] = 1;
  for (auto v : tree[u]) {
    if (v == pa) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    for (int j = 0; j < k; ++j) g[u][(j + 1) % k] += g[v][j];
  }
}
void dfs2(int u, int pa) {
  for (auto v : tree[u]) {
    if (v == pa) continue;
    int tmp[5] = {0};
    for (int i = 0; i < k; ++i)
      tmp[(i + 1) % k] = g[u][i] - g[v][(i - 1 + k) % k];
    for (int i = 0; i < k; ++i) g[v][i] += tmp[i];
    dfs2(v, u);
  }
}
long long solve(int n) {
  dfs1(1, 1);
  dfs2(1, 1);
  long long s[5] = {0};
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j < k; ++j) s[j] += g[i][j];
  for (int i = 1; i < k; ++i) s[i] /= 2;
  long long dis = 0, sum = 0;
  for (int i = 1; i <= n; ++i) dis += 1LL * siz[i] * (n - siz[i]);
  for (int i = 1; i < k; ++i) {
    dis -= 1LL * i * s[i];
    sum += s[i];
  }
  dis = dis / k + sum;
  return dis;
}
int main() {
  int n;
  scanf("%d %d", &n, &k);
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d %d", &u, &v);
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  printf("%lld\n", solve(n));
  return 0;
}

```