#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 2000 * 1000 * 1000;
const int MAXN = 200010;
template <typename T>
inline T sqr(T n) {
  return n * n;
}
int n, k;
vector<int> g[MAXN];
long long sum[MAXN][7];
long long dp[MAXN];
long long size[MAXN][7];
void dfs(int v, int par = -1) {
  size[v][0] = 1;
  for (int to : g[v]) {
    if (to != par) {
      dfs(to, v);
      for (int mod = 1; mod <= k; mod++) {
        size[v][mod % k] += size[to][mod - 1];
        sum[v][mod % k] += sum[to][mod - 1];
      }
      sum[v][0] += size[to][k - 1];
    }
  }
  for (int m2 = 0; m2 < k; m2++) {
    long long summ2 = 0, sizem2 = 0;
    for (int to : g[v]) {
      if (to != par) {
        summ2 += sum[to][m2];
        sizem2 += size[to][m2];
      }
    }
    for (int m1 = 0; m1 < k; m1++) {
      for (int to : g[v]) {
        if (to != par) {
          long long osize = sizem2 - size[to][m2];
          long long isize = size[to][m1];
          dp[v] += sum[to][m1] * osize + (summ2 - sum[to][m2]) * isize;
          dp[v] += isize * osize * ((m1 + m2 + 2 + k - 1) / k);
        }
      }
    }
  }
  for (int m = 0; m < k; m++) {
    dp[v] += 2 * (sum[v][m] + size[v][m] * (m != 0));
  }
  for (int to : g[v]) {
    if (to != par) {
      dp[v] += dp[to];
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  printf("%lld\n", dp[1] / 2);
  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 double EPS = 1e-9;
const double PI = acos(-1.0);
const int MOD = 1000 * 1000 * 1000 + 7;
const int INF = 2000 * 1000 * 1000;
const int MAXN = 200010;
template <typename T>
inline T sqr(T n) {
  return n * n;
}
int n, k;
vector<int> g[MAXN];
long long sum[MAXN][7];
long long dp[MAXN];
long long size[MAXN][7];
void dfs(int v, int par = -1) {
  size[v][0] = 1;
  for (int to : g[v]) {
    if (to != par) {
      dfs(to, v);
      for (int mod = 1; mod <= k; mod++) {
        size[v][mod % k] += size[to][mod - 1];
        sum[v][mod % k] += sum[to][mod - 1];
      }
      sum[v][0] += size[to][k - 1];
    }
  }
  for (int m2 = 0; m2 < k; m2++) {
    long long summ2 = 0, sizem2 = 0;
    for (int to : g[v]) {
      if (to != par) {
        summ2 += sum[to][m2];
        sizem2 += size[to][m2];
      }
    }
    for (int m1 = 0; m1 < k; m1++) {
      for (int to : g[v]) {
        if (to != par) {
          long long osize = sizem2 - size[to][m2];
          long long isize = size[to][m1];
          dp[v] += sum[to][m1] * osize + (summ2 - sum[to][m2]) * isize;
          dp[v] += isize * osize * ((m1 + m2 + 2 + k - 1) / k);
        }
      }
    }
  }
  for (int m = 0; m < k; m++) {
    dp[v] += 2 * (sum[v][m] + size[v][m] * (m != 0));
  }
  for (int to : g[v]) {
    if (to != par) {
      dp[v] += dp[to];
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  printf("%lld\n", dp[1] / 2);
  return 0;
}

```