#include <bits/stdc++.h>
using namespace std;
int const N = 2e5 + 41;
int const M = 5;
int n, K;
vector<int> e[N];
vector<long long> cnt[N][M], sum[N][M];
long long ans = 0;
long long tmp[M];
void read() {
  scanf("%d %d", &n, &K);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u;
    --v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
}
void dfs(int u, int p) {
  if (u == p) e[u].push_back(u);
  for (int i = 0; i < e[u].size(); ++i) {
    int to = e[u][i];
    if (to == p) {
      swap(e[u][i], e[u][0]);
      continue;
    }
    dfs(to, u);
  }
}
void rec(int u, int p) {
  for (int i = 0; i < K; ++i) {
    cnt[u][i].resize(e[u].size());
    sum[u][i].resize(e[u].size());
  }
  cnt[u][0][0] = 1;
  for (int i = 1; i < e[u].size(); ++i) {
    int to = e[u][i];
    rec(to, u);
    for (int kb = 0; kb < K; ++kb) {
      for (int kc = 0; kc < K; ++kc) {
        long long cb = cnt[u][kb][i - 1];
        long long cc = cnt[to][kc][e[to].size() - 1];
        long long sb = sum[u][kb][i - 1];
        long long sc = sum[to][kc][e[to].size() - 1];
        long long add = (kb + kc) / K;
        if ((kb + kc) % K) ++add;
        ans += cb * sc;
        ans += cc * sb;
        ans += add * cb * cc;
      }
    }
    for (int j = 0; j < K; ++j) {
      cnt[u][j][i] = cnt[u][j][i - 1] + cnt[to][j][e[to].size() - 1];
      sum[u][j][i] = sum[u][j][i - 1] + sum[to][j][e[to].size() - 1];
    }
  }
  for (int i = 0; i < K; ++i) {
    tmp[i] = sum[u][(i - 1 + K) % K][e[u].size() - 1];
    if (i == 0) {
      tmp[i] += cnt[u][(i - 1 + K) % K][e[u].size() - 1];
    }
  }
  for (int i = 0; i < K; ++i) {
    sum[u][i][e[u].size() - 1] = tmp[i];
  }
  for (int i = 0; i < K; ++i) {
    tmp[i] = cnt[u][(i - 1 + K) % K][e[u].size() - 1];
  }
  for (int i = 0; i < K; ++i) {
    cnt[u][i][e[u].size() - 1] = tmp[i];
  }
}
void solve() {
  dfs(0, 0);
  rec(0, 0);
  printf("%lld\n", ans);
}
int main() {
  read();
  solve();
  return 0;
}

### Prompt

Your task is to create 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;
int const N = 2e5 + 41;
int const M = 5;
int n, K;
vector<int> e[N];
vector<long long> cnt[N][M], sum[N][M];
long long ans = 0;
long long tmp[M];
void read() {
  scanf("%d %d", &n, &K);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u;
    --v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
}
void dfs(int u, int p) {
  if (u == p) e[u].push_back(u);
  for (int i = 0; i < e[u].size(); ++i) {
    int to = e[u][i];
    if (to == p) {
      swap(e[u][i], e[u][0]);
      continue;
    }
    dfs(to, u);
  }
}
void rec(int u, int p) {
  for (int i = 0; i < K; ++i) {
    cnt[u][i].resize(e[u].size());
    sum[u][i].resize(e[u].size());
  }
  cnt[u][0][0] = 1;
  for (int i = 1; i < e[u].size(); ++i) {
    int to = e[u][i];
    rec(to, u);
    for (int kb = 0; kb < K; ++kb) {
      for (int kc = 0; kc < K; ++kc) {
        long long cb = cnt[u][kb][i - 1];
        long long cc = cnt[to][kc][e[to].size() - 1];
        long long sb = sum[u][kb][i - 1];
        long long sc = sum[to][kc][e[to].size() - 1];
        long long add = (kb + kc) / K;
        if ((kb + kc) % K) ++add;
        ans += cb * sc;
        ans += cc * sb;
        ans += add * cb * cc;
      }
    }
    for (int j = 0; j < K; ++j) {
      cnt[u][j][i] = cnt[u][j][i - 1] + cnt[to][j][e[to].size() - 1];
      sum[u][j][i] = sum[u][j][i - 1] + sum[to][j][e[to].size() - 1];
    }
  }
  for (int i = 0; i < K; ++i) {
    tmp[i] = sum[u][(i - 1 + K) % K][e[u].size() - 1];
    if (i == 0) {
      tmp[i] += cnt[u][(i - 1 + K) % K][e[u].size() - 1];
    }
  }
  for (int i = 0; i < K; ++i) {
    sum[u][i][e[u].size() - 1] = tmp[i];
  }
  for (int i = 0; i < K; ++i) {
    tmp[i] = cnt[u][(i - 1 + K) % K][e[u].size() - 1];
  }
  for (int i = 0; i < K; ++i) {
    cnt[u][i][e[u].size() - 1] = tmp[i];
  }
}
void solve() {
  dfs(0, 0);
  rec(0, 0);
  printf("%lld\n", ans);
}
int main() {
  read();
  solve();
  return 0;
}

```