#include <bits/stdc++.h>
using namespace std;
const int MAXK = 5;
const int MAXN = 200000;
int n, k;
struct Path {
  Path() : num(), sj() {}
  int64_t num;
  int64_t sj;
};
struct PathK {
  Path p[MAXK];
  void addEdge() {
    Path t = p[k - 1];
    t.sj += t.num;
    for (int i = k - 1; i > 0; --i) p[i] = p[i - 1];
    p[0] = t;
    if (k < 2) {
      ++p[0].num;
      ++p[0].sj;
    } else {
      ++p[1].num;
    }
  }
  void add(const PathK& pk) {
    for (int i = 0; i < k; ++i) {
      p[i].num += pk.p[i].num;
      p[i].sj += pk.p[i].sj;
    }
  }
  void sub(const PathK& pk) {
    for (int i = 0; i < k; ++i) {
      p[i].num -= pk.p[i].num;
      p[i].sj -= pk.p[i].sj;
    }
  }
};
int64_t mul(const PathK& pk1, const PathK& pk2) {
  int64_t res = 0;
  for (int i1 = 0; i1 < k; ++i1)
    for (int i2 = 0; i2 < k; ++i2) {
      res += pk1.p[i1].num * pk2.p[i2].sj + pk1.p[i1].sj * pk2.p[i2].num;
      int64_t si = i1 + i2;
      if (si > 0) {
        int64_t mm = si > k ? 2 : 1;
        res += mm * (pk1.p[i1].num * pk2.p[i2].num);
      }
    }
  return res;
}
struct Vert {
  vector<int> chs;
  PathK np;
  void addCh(int c) {
    if (chs.size() == chs.capacity()) chs.reserve(chs.size() * 2);
    chs.push_back(c);
  }
};
vector<Vert> gr;
int64_t nRes;
void dfs(int fI, int vI) {
  Vert& v = gr[vI];
  int vchN = (int)v.chs.size();
  for (int i = 0; i < vchN; ++i) {
    int vchI = v.chs[i];
    if (vchI == fI) continue;
    dfs(vI, vchI);
    gr[vchI].np.addEdge();
    v.np.add(gr[vchI].np);
  }
  PathK tt;
  tt.p[0].num = 1;
  nRes += mul(tt, v.np);
  int64_t nResBtw = 0;
  for (int i = 0; i < vchN; ++i) {
    int vchI = v.chs[i];
    if (vchI == fI) continue;
    PathK ss = v.np;
    ss.sub(gr[vchI].np);
    nResBtw += mul(gr[vchI].np, ss);
  }
  nRes += nResBtw / 2;
}
int main() {
  cin >> n >> k;
  gr.resize(n);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    gr[a].addCh(b);
    gr[b].addCh(a);
  }
  dfs(-1, 0);
  cout << nRes;
  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;
const int MAXK = 5;
const int MAXN = 200000;
int n, k;
struct Path {
  Path() : num(), sj() {}
  int64_t num;
  int64_t sj;
};
struct PathK {
  Path p[MAXK];
  void addEdge() {
    Path t = p[k - 1];
    t.sj += t.num;
    for (int i = k - 1; i > 0; --i) p[i] = p[i - 1];
    p[0] = t;
    if (k < 2) {
      ++p[0].num;
      ++p[0].sj;
    } else {
      ++p[1].num;
    }
  }
  void add(const PathK& pk) {
    for (int i = 0; i < k; ++i) {
      p[i].num += pk.p[i].num;
      p[i].sj += pk.p[i].sj;
    }
  }
  void sub(const PathK& pk) {
    for (int i = 0; i < k; ++i) {
      p[i].num -= pk.p[i].num;
      p[i].sj -= pk.p[i].sj;
    }
  }
};
int64_t mul(const PathK& pk1, const PathK& pk2) {
  int64_t res = 0;
  for (int i1 = 0; i1 < k; ++i1)
    for (int i2 = 0; i2 < k; ++i2) {
      res += pk1.p[i1].num * pk2.p[i2].sj + pk1.p[i1].sj * pk2.p[i2].num;
      int64_t si = i1 + i2;
      if (si > 0) {
        int64_t mm = si > k ? 2 : 1;
        res += mm * (pk1.p[i1].num * pk2.p[i2].num);
      }
    }
  return res;
}
struct Vert {
  vector<int> chs;
  PathK np;
  void addCh(int c) {
    if (chs.size() == chs.capacity()) chs.reserve(chs.size() * 2);
    chs.push_back(c);
  }
};
vector<Vert> gr;
int64_t nRes;
void dfs(int fI, int vI) {
  Vert& v = gr[vI];
  int vchN = (int)v.chs.size();
  for (int i = 0; i < vchN; ++i) {
    int vchI = v.chs[i];
    if (vchI == fI) continue;
    dfs(vI, vchI);
    gr[vchI].np.addEdge();
    v.np.add(gr[vchI].np);
  }
  PathK tt;
  tt.p[0].num = 1;
  nRes += mul(tt, v.np);
  int64_t nResBtw = 0;
  for (int i = 0; i < vchN; ++i) {
    int vchI = v.chs[i];
    if (vchI == fI) continue;
    PathK ss = v.np;
    ss.sub(gr[vchI].np);
    nResBtw += mul(gr[vchI].np, ss);
  }
  nRes += nResBtw / 2;
}
int main() {
  cin >> n >> k;
  gr.resize(n);
  for (int i = 1; i < n; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    gr[a].addCh(b);
    gr[b].addCh(a);
  }
  dfs(-1, 0);
  cout << nRes;
  return 0;
}

```