#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long N = 2e5 + 5;
long long n, k;
long long ans = 0;
long long dp[N][5], tot[N], sz[N];
vector<long long> v[N];
void dfs(long long z, long long p) {
  sz[z] = 1;
  dp[z][0] = 1;
  reverse((v[z]).begin(), (v[z]).end());
  for (long long x : v[z]) {
    if (x == p) continue;
    dfs(x, z);
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        long long val = (i + j + k - 1) / k;
        ans += val * dp[x][(i + k - 1) % k] * (dp[z][j]);
      }
    }
    sz[z] += sz[x];
    for (long long i = 0; i < k; i++) {
      dp[z][(i + 1) % k] += dp[x][i];
    }
    tot[z] += tot[x] + dp[x][k - 1];
  }
  for (long long x : v[z]) {
    if (x == p) continue;
    ans += (sz[z] - sz[x]) * (tot[x] + dp[x][k - 1]);
  }
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  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;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T, typename T1>
T amax(T& a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T& a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long INF = 1e18;
const int32_t M = 1e9 + 7;
const int32_t MM = 998244353;
const long long N = 2e5 + 5;
long long n, k;
long long ans = 0;
long long dp[N][5], tot[N], sz[N];
vector<long long> v[N];
void dfs(long long z, long long p) {
  sz[z] = 1;
  dp[z][0] = 1;
  reverse((v[z]).begin(), (v[z]).end());
  for (long long x : v[z]) {
    if (x == p) continue;
    dfs(x, z);
    for (long long i = 0; i < k; i++) {
      for (long long j = 0; j < k; j++) {
        long long val = (i + j + k - 1) / k;
        ans += val * dp[x][(i + k - 1) % k] * (dp[z][j]);
      }
    }
    sz[z] += sz[x];
    for (long long i = 0; i < k; i++) {
      dp[z][(i + 1) % k] += dp[x][i];
    }
    tot[z] += tot[x] + dp[x][k - 1];
  }
  for (long long x : v[z]) {
    if (x == p) continue;
    ans += (sz[z] - sz[x]) * (tot[x] + dp[x][k - 1]);
  }
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  return 0;
}

```