#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
struct edge {
  int t, nxt;
} e[N << 1];
int n, K, cnt;
long long ans;
int head[N];
int f[N][10], g[N][10];
long long s[N], w[N];
void add(int u, int t) {
  e[++cnt].t = t;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs(int u, int fa) {
  f[u][0] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int t = e[i].t;
    if (t == fa) continue;
    dfs(t, u);
    s[u] += s[t] + f[t][0];
    for (int j = 0; j < K - 1; j++) f[u][j + 1] += f[t][j];
    f[u][0] += f[t][K - 1];
  }
  ans += s[u];
}
void dfs2(int u, int fa) {
  if (u != 1) {
    w[u] = w[fa] + g[fa][0];
    for (int j = 0; j < K - 1; j++) g[u][j + 1] += g[fa][j];
    g[u][0] += g[fa][K - 1];
    ans += w[u];
  }
  w[u] += s[u];
  for (int i = 0; i < K; i++) g[u][i] += f[u][i];
  for (int i = head[u]; i; i = e[i].nxt) {
    int t = e[i].t;
    if (t == fa) continue;
    w[u] -= s[t] + f[t][0];
    for (int j = 0; j < K - 1; j++) g[u][j + 1] -= f[t][j];
    g[u][0] -= f[t][K - 1];
    dfs2(t, u);
    w[u] += s[t] + f[t][0];
    for (int j = 0; j < K - 1; j++) g[u][j + 1] += f[t][j];
    g[u][0] += f[t][K - 1];
  }
}
signed main() {
  scanf("%d%d", &n, &K);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  dfs2(1, 0);
  printf("%lld\n", ans / 2);
}

### Prompt

Please formulate 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 N = 2e5 + 7;
struct edge {
  int t, nxt;
} e[N << 1];
int n, K, cnt;
long long ans;
int head[N];
int f[N][10], g[N][10];
long long s[N], w[N];
void add(int u, int t) {
  e[++cnt].t = t;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void dfs(int u, int fa) {
  f[u][0] = 1;
  for (int i = head[u]; i; i = e[i].nxt) {
    int t = e[i].t;
    if (t == fa) continue;
    dfs(t, u);
    s[u] += s[t] + f[t][0];
    for (int j = 0; j < K - 1; j++) f[u][j + 1] += f[t][j];
    f[u][0] += f[t][K - 1];
  }
  ans += s[u];
}
void dfs2(int u, int fa) {
  if (u != 1) {
    w[u] = w[fa] + g[fa][0];
    for (int j = 0; j < K - 1; j++) g[u][j + 1] += g[fa][j];
    g[u][0] += g[fa][K - 1];
    ans += w[u];
  }
  w[u] += s[u];
  for (int i = 0; i < K; i++) g[u][i] += f[u][i];
  for (int i = head[u]; i; i = e[i].nxt) {
    int t = e[i].t;
    if (t == fa) continue;
    w[u] -= s[t] + f[t][0];
    for (int j = 0; j < K - 1; j++) g[u][j + 1] -= f[t][j];
    g[u][0] -= f[t][K - 1];
    dfs2(t, u);
    w[u] += s[t] + f[t][0];
    for (int j = 0; j < K - 1; j++) g[u][j + 1] += f[t][j];
    g[u][0] += f[t][K - 1];
  }
}
signed main() {
  scanf("%d%d", &n, &K);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  dfs2(1, 0);
  printf("%lld\n", ans / 2);
}

```