#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
const int K = 10;
int n, k;
vector<int> g[N];
int maxp[N], siz[N], vis[N], rt;
void getRt(int u, int fa, int all) {
  siz[u] = 1, maxp[u] = 0;
  for (auto &v : g[u])
    if (v != fa && !vis[v]) {
      getRt(v, u, all);
      siz[u] += siz[v];
      maxp[u] = max(maxp[u], siz[v]);
    }
  maxp[u] = max(maxp[u], all - siz[u]);
  if (maxp[u] < maxp[rt]) rt = u;
}
long long sum[K], cnt[K], tsum[K], tcnt[K], ans;
void dfs2(int u, int fa, int d) {
  tsum[d % k] += (d + k - 1) / k, tcnt[d % k]++;
  for (auto &v : g[u])
    if (!vis[v] && v != fa) dfs2(v, u, d + 1);
}
void calc(int u) {
  for (int i = 0; i < k; i++) sum[i] = 0, cnt[i] = 0;
  cnt[0] = 1;
  for (auto &v : g[u])
    if (!vis[v]) {
      for (int i = 0; i < k; i++) tsum[i] = 0, tcnt[i] = 0;
      dfs2(v, u, 1);
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) {
          ans += cnt[i] * (tsum[j] - (j > 0) * tcnt[j]) +
                 (sum[i] - (i > 0) * cnt[i]) * tcnt[j];
          ans += (i + j + k - 1) / k * cnt[i] * tcnt[j];
        }
      for (int i = 0; i < k; i++) sum[i] += tsum[i], cnt[i] += tcnt[i];
    }
}
void dfs(int u) {
  vis[u] = 1;
  calc(u);
  for (auto &v : g[u])
    if (!vis[v]) {
      maxp[rt = 0] = N;
      getRt(v, 0, siz[v]);
      dfs(rt);
    }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  maxp[rt = 0] = N;
  getRt(1, 0, N);
  dfs(rt);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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 + 10;
const int K = 10;
int n, k;
vector<int> g[N];
int maxp[N], siz[N], vis[N], rt;
void getRt(int u, int fa, int all) {
  siz[u] = 1, maxp[u] = 0;
  for (auto &v : g[u])
    if (v != fa && !vis[v]) {
      getRt(v, u, all);
      siz[u] += siz[v];
      maxp[u] = max(maxp[u], siz[v]);
    }
  maxp[u] = max(maxp[u], all - siz[u]);
  if (maxp[u] < maxp[rt]) rt = u;
}
long long sum[K], cnt[K], tsum[K], tcnt[K], ans;
void dfs2(int u, int fa, int d) {
  tsum[d % k] += (d + k - 1) / k, tcnt[d % k]++;
  for (auto &v : g[u])
    if (!vis[v] && v != fa) dfs2(v, u, d + 1);
}
void calc(int u) {
  for (int i = 0; i < k; i++) sum[i] = 0, cnt[i] = 0;
  cnt[0] = 1;
  for (auto &v : g[u])
    if (!vis[v]) {
      for (int i = 0; i < k; i++) tsum[i] = 0, tcnt[i] = 0;
      dfs2(v, u, 1);
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) {
          ans += cnt[i] * (tsum[j] - (j > 0) * tcnt[j]) +
                 (sum[i] - (i > 0) * cnt[i]) * tcnt[j];
          ans += (i + j + k - 1) / k * cnt[i] * tcnt[j];
        }
      for (int i = 0; i < k; i++) sum[i] += tsum[i], cnt[i] += tcnt[i];
    }
}
void dfs(int u) {
  vis[u] = 1;
  calc(u);
  for (auto &v : g[u])
    if (!vis[v]) {
      maxp[rt = 0] = N;
      getRt(v, 0, siz[v]);
      dfs(rt);
    }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  maxp[rt = 0] = N;
  getRt(1, 0, N);
  dfs(rt);
  printf("%lld\n", ans);
  return 0;
}

```