#include <bits/stdc++.h>
using namespace std;
inline void Rd(int &res) {
  char c;
  res = 0;
  while (c = getchar(), c < '0')
    ;
  do {
    res = (res << 1) + (res << 3) + (c ^ 48);
  } while (c = getchar(), c >= '0');
}
const int M = 200005;
struct Node {
  int to, nxt;
} Edge[M << 1];
int Head[M], tot;
int n, k;
long long ans, cnt[M];
int depth[M];
map<int, int> Mp[M];
inline void Addedge(int a, int b) {
  Edge[tot] = (Node){b, Head[a]};
  Head[a] = tot++;
  Edge[tot] = (Node){a, Head[b]};
  Head[b] = tot++;
}
void work(int fr, int to, int x) {
  int sum1 = 0, sum2 = 0;
  for (map<int, int>::iterator it = Mp[fr].begin(); it != Mp[fr].end(); it++)
    sum1 += (it->second);
  for (map<int, int>::iterator it = Mp[to].begin(); it != Mp[to].end(); it++)
    sum2 += (it->second);
  ans += 1LL * sum1 * cnt[to] + 1LL * sum2 * cnt[fr];
  for (map<int, int>::iterator it1 = Mp[fr].begin(); it1 != Mp[fr].end(); it1++)
    for (map<int, int>::iterator it2 = Mp[to].begin(); it2 != Mp[to].end();
         it2++) {
      int d1 = it1->first, d2 = it2->first;
      int cnt1 = it1->second, cnt2 = it2->second;
      int dist = d1 + d2 - 2 * depth[x];
      ans += 1LL * ((dist - 1) / k + 1) * cnt1 * cnt2;
    }
}
void add(int fr, int to) {
  cnt[to] += cnt[fr];
  for (map<int, int>::iterator it = Mp[fr].begin(); it != Mp[fr].end(); it++)
    Mp[to][it->first] += (it->second);
}
void del(int x) {
  map<int, int> T;
  for (map<int, int>::iterator it = Mp[x].begin(); it != Mp[x].end(); it++) {
    int d = it->first;
    if (d - k > depth[x]) {
      T[d - k] += (it->second);
      cnt[x] += (it->second);
    } else
      T[d] += (it->second);
  }
  Mp[x] = T;
}
void solve(int x, int f) {
  int sz = 0;
  for (int i = Head[x]; ~i; i = Edge[i].nxt) {
    int to = Edge[i].to;
    if (to == f) continue;
    sz++;
    depth[to] = depth[x] + 1;
    solve(to, x);
  }
  Mp[x][depth[x]] = 1;
  if (sz) {
    int fi;
    for (int i = Head[x]; ~i; i = Edge[i].nxt) {
      int to = Edge[i].to;
      if (to == f) continue;
      fi = to;
      break;
    }
    for (int i = Head[x]; ~i; i = Edge[i].nxt) {
      int to = Edge[i].to;
      if (to == f || to == fi) continue;
      work(to, fi, x);
      add(to, fi);
    }
    ans += cnt[fi];
    for (map<int, int>::iterator it = Mp[fi].begin(); it != Mp[fi].end();
         it++) {
      int dist = (it->first) - depth[x];
      ans += 1LL * ((dist - 1) / k + 1) * (it->second);
    }
    cnt[x] += cnt[fi];
    for (map<int, int>::iterator it = Mp[fi].begin(); it != Mp[fi].end(); it++)
      Mp[x][it->first] += (it->second);
    del(x);
  }
}
int main() {
  Rd(n), Rd(k);
  memset(Head, -1, sizeof(Head));
  for (int i = 1; i < n; i++) {
    int a, b;
    Rd(a), Rd(b);
    Addedge(a, b);
  }
  depth[1] = 1;
  solve(1, 0);
  cout << ans << endl;
  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;
inline void Rd(int &res) {
  char c;
  res = 0;
  while (c = getchar(), c < '0')
    ;
  do {
    res = (res << 1) + (res << 3) + (c ^ 48);
  } while (c = getchar(), c >= '0');
}
const int M = 200005;
struct Node {
  int to, nxt;
} Edge[M << 1];
int Head[M], tot;
int n, k;
long long ans, cnt[M];
int depth[M];
map<int, int> Mp[M];
inline void Addedge(int a, int b) {
  Edge[tot] = (Node){b, Head[a]};
  Head[a] = tot++;
  Edge[tot] = (Node){a, Head[b]};
  Head[b] = tot++;
}
void work(int fr, int to, int x) {
  int sum1 = 0, sum2 = 0;
  for (map<int, int>::iterator it = Mp[fr].begin(); it != Mp[fr].end(); it++)
    sum1 += (it->second);
  for (map<int, int>::iterator it = Mp[to].begin(); it != Mp[to].end(); it++)
    sum2 += (it->second);
  ans += 1LL * sum1 * cnt[to] + 1LL * sum2 * cnt[fr];
  for (map<int, int>::iterator it1 = Mp[fr].begin(); it1 != Mp[fr].end(); it1++)
    for (map<int, int>::iterator it2 = Mp[to].begin(); it2 != Mp[to].end();
         it2++) {
      int d1 = it1->first, d2 = it2->first;
      int cnt1 = it1->second, cnt2 = it2->second;
      int dist = d1 + d2 - 2 * depth[x];
      ans += 1LL * ((dist - 1) / k + 1) * cnt1 * cnt2;
    }
}
void add(int fr, int to) {
  cnt[to] += cnt[fr];
  for (map<int, int>::iterator it = Mp[fr].begin(); it != Mp[fr].end(); it++)
    Mp[to][it->first] += (it->second);
}
void del(int x) {
  map<int, int> T;
  for (map<int, int>::iterator it = Mp[x].begin(); it != Mp[x].end(); it++) {
    int d = it->first;
    if (d - k > depth[x]) {
      T[d - k] += (it->second);
      cnt[x] += (it->second);
    } else
      T[d] += (it->second);
  }
  Mp[x] = T;
}
void solve(int x, int f) {
  int sz = 0;
  for (int i = Head[x]; ~i; i = Edge[i].nxt) {
    int to = Edge[i].to;
    if (to == f) continue;
    sz++;
    depth[to] = depth[x] + 1;
    solve(to, x);
  }
  Mp[x][depth[x]] = 1;
  if (sz) {
    int fi;
    for (int i = Head[x]; ~i; i = Edge[i].nxt) {
      int to = Edge[i].to;
      if (to == f) continue;
      fi = to;
      break;
    }
    for (int i = Head[x]; ~i; i = Edge[i].nxt) {
      int to = Edge[i].to;
      if (to == f || to == fi) continue;
      work(to, fi, x);
      add(to, fi);
    }
    ans += cnt[fi];
    for (map<int, int>::iterator it = Mp[fi].begin(); it != Mp[fi].end();
         it++) {
      int dist = (it->first) - depth[x];
      ans += 1LL * ((dist - 1) / k + 1) * (it->second);
    }
    cnt[x] += cnt[fi];
    for (map<int, int>::iterator it = Mp[fi].begin(); it != Mp[fi].end(); it++)
      Mp[x][it->first] += (it->second);
    del(x);
  }
}
int main() {
  Rd(n), Rd(k);
  memset(Head, -1, sizeof(Head));
  for (int i = 1; i < n; i++) {
    int a, b;
    Rd(a), Rd(b);
    Addedge(a, b);
  }
  depth[1] = 1;
  solve(1, 0);
  cout << ans << endl;
  return 0;
}

```