#include <bits/stdc++.h>
using namespace std;
const int N = 2e6;
struct Node {
  long long cnt, dp[5];
};
long long n, k;
vector<int> g[N];
Node tr[N];
long long ans;
void build(int c, int p) {
  auto& nn(tr[c]);
  nn.cnt = nn.dp[0] = 1;
  for (auto it : g[c]) {
    if (it != p) {
      build(it, c);
      nn.cnt += tr[it].cnt;
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) {
          ans += nn.dp[i] * tr[it].dp[j] * ((((k - (i + j + 1)) % k) + k) % k);
        }
      for (int i = 0; i < k; i++) nn.dp[(i + 1) % k] += tr[it].dp[i];
    }
  }
  ans += nn.cnt * (n - nn.cnt);
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  build(1, 0);
  printf("%lld\n", ans / k);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 = 2e6;
struct Node {
  long long cnt, dp[5];
};
long long n, k;
vector<int> g[N];
Node tr[N];
long long ans;
void build(int c, int p) {
  auto& nn(tr[c]);
  nn.cnt = nn.dp[0] = 1;
  for (auto it : g[c]) {
    if (it != p) {
      build(it, c);
      nn.cnt += tr[it].cnt;
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++) {
          ans += nn.dp[i] * tr[it].dp[j] * ((((k - (i + j + 1)) % k) + k) % k);
        }
      for (int i = 0; i < k; i++) nn.dp[(i + 1) % k] += tr[it].dp[i];
    }
  }
  ans += nn.cnt * (n - nn.cnt);
}
int main() {
  scanf("%lld%lld", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    g[b].push_back(a);
  }
  build(1, 0);
  printf("%lld\n", ans / k);
  return 0;
}

```