#include <bits/stdc++.h>
const int INF = 0x7F7F7F7F;
const double EPS = 1e-10;
const long long mod7 = 1e9 + 7;
const long long mod9 = 1e9 + 9;
using namespace std;
inline long long rit() {
  long long f = 0, key = 1;
  char c;
  do {
    c = getchar();
    if (c == '-') key = -1;
  } while (c < '0' || c > '9');
  do {
    f = f * 10 + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  return f * key;
}
inline void fprt(double f) { printf("%.08lf", f); }
void init() {}
const int Z = 200005;
vector<int> v[Z];
int n, k;
long long dp[Z], size[Z], dp2[Z], cnt[Z][5], cnt2[Z][5], mem[Z];
long long ans;
void dfs(int p, int fa) {
  size[p] = 1;
  for (auto f : v[p]) {
    if (f == fa) continue;
    dfs(f, p);
    size[p] += size[f];
    dp[p] += dp[f] + size[f];
    for (int i = 0; i < k; ++i) cnt[p][i] += cnt[f][(i + k - 1) % k];
  }
  cnt[p][0]++;
}
void dfs2(int p, int fa) {
  if (p == 1) {
    dp2[p] = 0;
    for (int i = 0; i < k; ++i) cnt2[p][i] = cnt[p][i];
  } else {
    dp2[p] = dp2[fa] + (dp[fa] - dp[p] - size[p]) + (n - size[p]);
    for (int i = 0; i < k; ++i) {
      cnt2[p][i] =
          cnt[p][i] + (cnt2[fa][(i + k - 1) % k] - cnt[p][(i + k * 2 - 2) % k]);
    }
  }
  for (auto f : v[p]) {
    if (f == fa) continue;
    dfs2(f, p);
  }
  long long sum = dp2[p] + dp[p];
  for (int i = 1; i < k; ++i) {
    sum -= cnt2[p][i] * i;
  }
  mem[p] += sum / k;
  for (int i = 1; i < k; ++i) mem[p] += cnt2[p][i];
  ans += mem[p];
  return;
}
void read() {
  int a, b;
  n = rit(), k = rit();
  for (int i = 0; i < n - 1; ++i) {
    a = rit(), b = rit();
    v[a].push_back(b);
    v[b].push_back(a);
  }
}
void solve() {
  dfs(1, 0);
  dfs2(1, 0);
  cout << ans / 2 << endl;
}
int main() {
  int nn = 1;
  while (nn--) {
    init(), read(), solve();
  }
  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>
const int INF = 0x7F7F7F7F;
const double EPS = 1e-10;
const long long mod7 = 1e9 + 7;
const long long mod9 = 1e9 + 9;
using namespace std;
inline long long rit() {
  long long f = 0, key = 1;
  char c;
  do {
    c = getchar();
    if (c == '-') key = -1;
  } while (c < '0' || c > '9');
  do {
    f = f * 10 + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  return f * key;
}
inline void fprt(double f) { printf("%.08lf", f); }
void init() {}
const int Z = 200005;
vector<int> v[Z];
int n, k;
long long dp[Z], size[Z], dp2[Z], cnt[Z][5], cnt2[Z][5], mem[Z];
long long ans;
void dfs(int p, int fa) {
  size[p] = 1;
  for (auto f : v[p]) {
    if (f == fa) continue;
    dfs(f, p);
    size[p] += size[f];
    dp[p] += dp[f] + size[f];
    for (int i = 0; i < k; ++i) cnt[p][i] += cnt[f][(i + k - 1) % k];
  }
  cnt[p][0]++;
}
void dfs2(int p, int fa) {
  if (p == 1) {
    dp2[p] = 0;
    for (int i = 0; i < k; ++i) cnt2[p][i] = cnt[p][i];
  } else {
    dp2[p] = dp2[fa] + (dp[fa] - dp[p] - size[p]) + (n - size[p]);
    for (int i = 0; i < k; ++i) {
      cnt2[p][i] =
          cnt[p][i] + (cnt2[fa][(i + k - 1) % k] - cnt[p][(i + k * 2 - 2) % k]);
    }
  }
  for (auto f : v[p]) {
    if (f == fa) continue;
    dfs2(f, p);
  }
  long long sum = dp2[p] + dp[p];
  for (int i = 1; i < k; ++i) {
    sum -= cnt2[p][i] * i;
  }
  mem[p] += sum / k;
  for (int i = 1; i < k; ++i) mem[p] += cnt2[p][i];
  ans += mem[p];
  return;
}
void read() {
  int a, b;
  n = rit(), k = rit();
  for (int i = 0; i < n - 1; ++i) {
    a = rit(), b = rit();
    v[a].push_back(b);
    v[b].push_back(a);
  }
}
void solve() {
  dfs(1, 0);
  dfs2(1, 0);
  cout << ans / 2 << endl;
}
int main() {
  int nn = 1;
  while (nn--) {
    init(), read(), solve();
  }
  return 0;
}

```