#include <bits/stdc++.h>
using namespace std;
inline void Max(int &x, int t) {
  if (t > x) {
    x = t;
  }
}
inline void Min(int &x, int t) {
  if (t < x) {
    x = t;
  }
}
const int maxn = 200000 + 9;
long long ans = 0, tot[maxn], cnt[maxn][6], cnt0[6];
int N, K;
vector<int> g[maxn];
void dfs(int x, int fa) {
  cnt[x][0] = 1;
  for (auto e : g[x])
    if (e != fa) {
      dfs(e, x);
      for (int i = 0; i < K; i++) {
        cnt[x][(i + 1) % K] += cnt[e][i];
      }
      tot[x] += tot[e] + cnt[e][0];
    }
}
void dfs0(int x, int fa) {
  ans += tot[x];
  for (auto e : g[x])
    if (e != fa) {
      for (int i = 0; i < K; i++) {
        cnt0[i] = cnt[e][i];
      }
      for (int i = 0; i < K; i++) {
        cnt[e][(i + 1) % K] += cnt[x][i] - cnt0[(i + K - 1) % K];
      }
      tot[e] += tot[x] - tot[e] - cnt0[0] + cnt[x][0] - cnt0[K - 1];
      dfs0(e, x);
    }
}
int main() {
  cin >> N >> K;
  for (int i = 1; i < N; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(cnt, 0, sizeof cnt);
  memset(tot, 0, sizeof tot);
  dfs(0, -1);
  dfs0(0, -1);
  cout << ans / 2 << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
inline void Max(int &x, int t) {
  if (t > x) {
    x = t;
  }
}
inline void Min(int &x, int t) {
  if (t < x) {
    x = t;
  }
}
const int maxn = 200000 + 9;
long long ans = 0, tot[maxn], cnt[maxn][6], cnt0[6];
int N, K;
vector<int> g[maxn];
void dfs(int x, int fa) {
  cnt[x][0] = 1;
  for (auto e : g[x])
    if (e != fa) {
      dfs(e, x);
      for (int i = 0; i < K; i++) {
        cnt[x][(i + 1) % K] += cnt[e][i];
      }
      tot[x] += tot[e] + cnt[e][0];
    }
}
void dfs0(int x, int fa) {
  ans += tot[x];
  for (auto e : g[x])
    if (e != fa) {
      for (int i = 0; i < K; i++) {
        cnt0[i] = cnt[e][i];
      }
      for (int i = 0; i < K; i++) {
        cnt[e][(i + 1) % K] += cnt[x][i] - cnt0[(i + K - 1) % K];
      }
      tot[e] += tot[x] - tot[e] - cnt0[0] + cnt[x][0] - cnt0[K - 1];
      dfs0(e, x);
    }
}
int main() {
  cin >> N >> K;
  for (int i = 1; i < N; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(cnt, 0, sizeof cnt);
  memset(tot, 0, sizeof tot);
  dfs(0, -1);
  dfs0(0, -1);
  cout << ans / 2 << endl;
  return 0;
}

```