#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > tree;
int n, k;
long long ans;
vector<int> num;
vector<vector<int> > numST;
void preprocess(void) { return; }
void MODdfs(int v, int p, int h) {
  num[v] = 1;
  numST[v][h % k] = 1;
  for (int i = 0; i < (int)tree[v].size(); i++) {
    int u = tree[v][i];
    if (u != p) {
      MODdfs(u, v, h + 1);
      num[v] += num[u];
      for (int x = 0; x < k; x++) {
        for (int y = 0; y < k; y++) {
          int dist = (x + y - ((2 * h) % k)) + k;
          dist %= k;
          int tbm = k - dist;
          tbm %= k;
          ans += ((long long)tbm) * numST[v][x] * numST[u][y];
        }
      }
      for (int j = 0; j < k; j++) {
        numST[v][j] += numST[u][j];
      }
    }
  }
  ans += ((long long)num[v]) * (n - num[v]);
}
int main(void) {
  preprocess();
  int t;
  t = 1;
  for (int z = 1; z <= t; z++) {
    scanf("%d", &n), scanf("%d", &k);
    tree.resize(n);
    num.assign(n, 0);
    numST.assign(n, vector<int>(k, 0));
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      scanf("%d", &a), scanf("%d", &b);
      a--, b--;
      tree[a].push_back(b);
      tree[b].push_back(a);
    }
    MODdfs(0, -1, 1);
    ans /= k;
    printf("%lld\n", ans);
  }
  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;
vector<vector<int> > tree;
int n, k;
long long ans;
vector<int> num;
vector<vector<int> > numST;
void preprocess(void) { return; }
void MODdfs(int v, int p, int h) {
  num[v] = 1;
  numST[v][h % k] = 1;
  for (int i = 0; i < (int)tree[v].size(); i++) {
    int u = tree[v][i];
    if (u != p) {
      MODdfs(u, v, h + 1);
      num[v] += num[u];
      for (int x = 0; x < k; x++) {
        for (int y = 0; y < k; y++) {
          int dist = (x + y - ((2 * h) % k)) + k;
          dist %= k;
          int tbm = k - dist;
          tbm %= k;
          ans += ((long long)tbm) * numST[v][x] * numST[u][y];
        }
      }
      for (int j = 0; j < k; j++) {
        numST[v][j] += numST[u][j];
      }
    }
  }
  ans += ((long long)num[v]) * (n - num[v]);
}
int main(void) {
  preprocess();
  int t;
  t = 1;
  for (int z = 1; z <= t; z++) {
    scanf("%d", &n), scanf("%d", &k);
    tree.resize(n);
    num.assign(n, 0);
    numST.assign(n, vector<int>(k, 0));
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      scanf("%d", &a), scanf("%d", &b);
      a--, b--;
      tree[a].push_back(b);
      tree[b].push_back(a);
    }
    MODdfs(0, -1, 1);
    ans /= k;
    printf("%lld\n", ans);
  }
  return 0;
}

```