#include <bits/stdc++.h>
using namespace std;
namespace READ {
inline int nxt() {
  int n;
  scanf("%d", &n);
  return n;
}
inline long long lnxt() {
  long long n;
  scanf("%I64d", &n);
  return n;
}
const int inf = 1000 * 1000 * 1000;
const long long linf = (long long)inf * (long long)inf;
const long long mod = 1000 * 1000 * 1000 + 7;
const int num = 200 * 1000 + 7;
const int N = 356;
const long double PI = acos((long double)-1.0);
}  // namespace READ
using namespace READ;
int n, k;
vector<int> g[num];
long long qan[num][7];
long long dp[num][7];
long long ans = 0;
void dfs(int v, int p) {
  for (int to : g[v]) {
    if (to == p) continue;
    dfs(to, v);
  }
  qan[v][0]++;
  for (auto to : g[v]) {
    if (to == p) continue;
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j) {
        ans += dp[to][j] * qan[v][i] + qan[to][j] * dp[v][i];
        ans += qan[v][i] * qan[to][j] * (1 + (i + j) / k);
      }
    for (int i = 0; i < k; ++i) {
      dp[v][(i + 1) % k] += dp[to][i] + (i == k - 1 ? qan[to][i] : 0);
      qan[v][(i + 1) % k] += qan[to][i];
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u = nxt(), v = nxt();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 1);
  cout << ans << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
namespace READ {
inline int nxt() {
  int n;
  scanf("%d", &n);
  return n;
}
inline long long lnxt() {
  long long n;
  scanf("%I64d", &n);
  return n;
}
const int inf = 1000 * 1000 * 1000;
const long long linf = (long long)inf * (long long)inf;
const long long mod = 1000 * 1000 * 1000 + 7;
const int num = 200 * 1000 + 7;
const int N = 356;
const long double PI = acos((long double)-1.0);
}  // namespace READ
using namespace READ;
int n, k;
vector<int> g[num];
long long qan[num][7];
long long dp[num][7];
long long ans = 0;
void dfs(int v, int p) {
  for (int to : g[v]) {
    if (to == p) continue;
    dfs(to, v);
  }
  qan[v][0]++;
  for (auto to : g[v]) {
    if (to == p) continue;
    for (int i = 0; i < k; ++i)
      for (int j = 0; j < k; ++j) {
        ans += dp[to][j] * qan[v][i] + qan[to][j] * dp[v][i];
        ans += qan[v][i] * qan[to][j] * (1 + (i + j) / k);
      }
    for (int i = 0; i < k; ++i) {
      dp[v][(i + 1) % k] += dp[to][i] + (i == k - 1 ? qan[to][i] : 0);
      qan[v][(i + 1) % k] += qan[to][i];
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u = nxt(), v = nxt();
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 1);
  cout << ans << endl;
  return 0;
}

```