#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long maxn = 3e5;
const long long maxk = 6;
long long n, m, k;
long long D[maxn][maxk];
vector<long long> V[maxn];
long long H[maxn];
long long ans = 0;
void dfs(long long v, long long p) {
  long long q, w;
  D[v][0]++;
  if (v != 0) {
    H[v] = H[p] + 1;
  }
  for (q = 0; q < V[v].size(); q++) {
    if (V[v][q] == p) {
      continue;
    }
    dfs(V[v][q], v);
    for (w = 0; w < k; w++) {
      D[v][w] += D[V[v][q]][(w - 1 + k) % k];
    }
  }
}
void dfs1(long long v, long long p, long long curans, vector<long long> down,
          vector<long long> up) {
  long long q, w;
  ans += curans;
  for (q = 0; q < V[v].size(); q++) {
    if (V[v][q] == p) {
      continue;
    }
    long long v1 = V[v][q];
    vector<long long> up2, down2;
    up2 = up;
    down2 = down;
    for (w = 0; w < k; w++) {
      down2[w] -= (D[v][w] - D[v1][(w - 1 + k) % k]);
      up2[w] += (D[v][w] - D[v1][(w - 1 + k) % k]);
    }
    vector<long long> up1, down1;
    up1.resize(k);
    down1.resize(k);
    for (w = 0; w < k; w++) {
      up1[w] = up2[(w - 1 + k) % k];
      down1[w] = down2[(w + 1) % k];
    }
    long long curans1 = curans + (up1[1 % k] - down1[0]);
    dfs1(V[v][q], v, curans1, down1, up1);
  }
}
int main() {
  long long q, w, e, t, a, b, c;
  cin >> n >> k;
  for (q = 0; q < n - 1; q++) {
    cin >> a >> b;
    a--;
    b--;
    V[a].push_back(b);
    V[b].push_back(a);
  }
  dfs(0, 0);
  long long curans = 0;
  vector<long long> down, up;
  down.resize(k);
  up.resize(k);
  for (q = 0; q < k; q++) {
    up[q] = 0;
    down[q] = D[0][q];
  }
  for (q = 0; q < n; q++) {
    curans += (H[q] + k - 1) / k;
  }
  dfs1(0, 0, curans, down, up);
  cout << ans / 2;
  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>
#pragma GCC optimize("O3")
using namespace std;
const long long maxn = 3e5;
const long long maxk = 6;
long long n, m, k;
long long D[maxn][maxk];
vector<long long> V[maxn];
long long H[maxn];
long long ans = 0;
void dfs(long long v, long long p) {
  long long q, w;
  D[v][0]++;
  if (v != 0) {
    H[v] = H[p] + 1;
  }
  for (q = 0; q < V[v].size(); q++) {
    if (V[v][q] == p) {
      continue;
    }
    dfs(V[v][q], v);
    for (w = 0; w < k; w++) {
      D[v][w] += D[V[v][q]][(w - 1 + k) % k];
    }
  }
}
void dfs1(long long v, long long p, long long curans, vector<long long> down,
          vector<long long> up) {
  long long q, w;
  ans += curans;
  for (q = 0; q < V[v].size(); q++) {
    if (V[v][q] == p) {
      continue;
    }
    long long v1 = V[v][q];
    vector<long long> up2, down2;
    up2 = up;
    down2 = down;
    for (w = 0; w < k; w++) {
      down2[w] -= (D[v][w] - D[v1][(w - 1 + k) % k]);
      up2[w] += (D[v][w] - D[v1][(w - 1 + k) % k]);
    }
    vector<long long> up1, down1;
    up1.resize(k);
    down1.resize(k);
    for (w = 0; w < k; w++) {
      up1[w] = up2[(w - 1 + k) % k];
      down1[w] = down2[(w + 1) % k];
    }
    long long curans1 = curans + (up1[1 % k] - down1[0]);
    dfs1(V[v][q], v, curans1, down1, up1);
  }
}
int main() {
  long long q, w, e, t, a, b, c;
  cin >> n >> k;
  for (q = 0; q < n - 1; q++) {
    cin >> a >> b;
    a--;
    b--;
    V[a].push_back(b);
    V[b].push_back(a);
  }
  dfs(0, 0);
  long long curans = 0;
  vector<long long> down, up;
  down.resize(k);
  up.resize(k);
  for (q = 0; q < k; q++) {
    up[q] = 0;
    down[q] = D[0][q];
  }
  for (q = 0; q < n; q++) {
    curans += (H[q] + k - 1) / k;
  }
  dfs1(0, 0, curans, down, up);
  cout << ans / 2;
  return 0;
}

```