#include <bits/stdc++.h>
using namespace std;
const int N = 200001;
const int K = 6;
long long dp[N][K];
long long cp[N][K];
vector<int> g[N];
long long ans = 0;
void dfs(int x, int p, int k) {
  for (auto& y : g[x]) {
    if (y == p) continue;
    dfs(y, x, k);
    for (int i = 0; i < k; ++i) {
      dp[x][i] += dp[y][(k + i - 1) % k];
    }
    dp[x][1 % k] += 1;
    dp[x][k] += dp[y][0] + dp[y][k] + 1;
  }
}
void dfs2(int x, int p, int k) {
  ans += dp[x][k] + cp[x][k];
  for (auto& y : g[x]) {
    if (y == p) continue;
    for (int i = 0; i < k; ++i) {
      cp[y][i] += dp[x][(k + i - 1) % k] - dp[y][(k + k + i - 2) % k];
      cp[y][i] += cp[x][(k + i - 1) % k];
    }
    cp[y][2 % k] -= 1;
    cp[y][1 % k] += 1;
    cp[y][k] += cp[x][0] + cp[x][k];
    cp[y][k] += (dp[x][0] - dp[y][k - 1]) + (dp[x][k] - dp[y][0] - dp[y][k]);
    if (k == 1) cp[y][k] -= 1;
    dfs2(y, x, k);
  }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0, k);
  dfs2(1, 0, k);
  cout << ans / 2 << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 int N = 200001;
const int K = 6;
long long dp[N][K];
long long cp[N][K];
vector<int> g[N];
long long ans = 0;
void dfs(int x, int p, int k) {
  for (auto& y : g[x]) {
    if (y == p) continue;
    dfs(y, x, k);
    for (int i = 0; i < k; ++i) {
      dp[x][i] += dp[y][(k + i - 1) % k];
    }
    dp[x][1 % k] += 1;
    dp[x][k] += dp[y][0] + dp[y][k] + 1;
  }
}
void dfs2(int x, int p, int k) {
  ans += dp[x][k] + cp[x][k];
  for (auto& y : g[x]) {
    if (y == p) continue;
    for (int i = 0; i < k; ++i) {
      cp[y][i] += dp[x][(k + i - 1) % k] - dp[y][(k + k + i - 2) % k];
      cp[y][i] += cp[x][(k + i - 1) % k];
    }
    cp[y][2 % k] -= 1;
    cp[y][1 % k] += 1;
    cp[y][k] += cp[x][0] + cp[x][k];
    cp[y][k] += (dp[x][0] - dp[y][k - 1]) + (dp[x][k] - dp[y][0] - dp[y][k]);
    if (k == 1) cp[y][k] -= 1;
    dfs2(y, x, k);
  }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0, k);
  dfs2(1, 0, k);
  cout << ans / 2 << endl;
  return 0;
}

```