#include <bits/stdc++.h>
using namespace std;
template <typename T>
bool chmax(T& a, const T& b) {
  return a < b ? (a = b, true) : false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  return b < a ? (a = b, true) : false;
}
int N, K;
vector<int> edge[200001];
vector<long long> dp[200001], dp2[200001];
vector<long long> dfs(int curr, int prev) {
  dp[curr].assign(K + 1, 0);
  vector<long long>& res = dp[curr];
  for (const auto& it : edge[curr])
    if (it != prev) {
      vector<long long> v = dfs(it, curr);
      for (int i = (int)(0); i < (int)(K); i++) {
        res[(i + 1) % K] += v[i];
      }
      res[K] += v[K] + v[0];
    }
  res[0]++;
  return res;
}
void dfs2(int curr, int prev) {
  dp2[curr].assign(K + 1, 0);
  vector<long long>& res = dp2[curr];
  if (curr != 1) {
    for (int i = (int)(0); i < (int)(K); i++) {
      res[(i + 1) % K] += dp[prev][i];
      res[(i + 1) % K] += dp2[prev][i];
      res[(i + 2) % K] -= dp[curr][i];
    }
    res[K] += dp[prev][K] + dp2[prev][K] - dp[curr][K];
    res[K] += dp[prev][0] + dp2[prev][0] - dp[curr][0] - dp[curr][K - 1];
  }
  for (const auto& it : edge[curr])
    if (it != prev) {
      dfs2(it, curr);
    }
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = (int)(0); i < (int)(N - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs(1, -1);
  dfs2(1, -1);
  long long ans = 0;
  for (int i = (int)(1); i < (int)(N + 1); i++) {
    ans += dp[i][K] + dp2[i][K];
  }
  ans /= 2;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
template <typename T>
bool chmax(T& a, const T& b) {
  return a < b ? (a = b, true) : false;
}
template <typename T>
bool chmin(T& a, const T& b) {
  return b < a ? (a = b, true) : false;
}
int N, K;
vector<int> edge[200001];
vector<long long> dp[200001], dp2[200001];
vector<long long> dfs(int curr, int prev) {
  dp[curr].assign(K + 1, 0);
  vector<long long>& res = dp[curr];
  for (const auto& it : edge[curr])
    if (it != prev) {
      vector<long long> v = dfs(it, curr);
      for (int i = (int)(0); i < (int)(K); i++) {
        res[(i + 1) % K] += v[i];
      }
      res[K] += v[K] + v[0];
    }
  res[0]++;
  return res;
}
void dfs2(int curr, int prev) {
  dp2[curr].assign(K + 1, 0);
  vector<long long>& res = dp2[curr];
  if (curr != 1) {
    for (int i = (int)(0); i < (int)(K); i++) {
      res[(i + 1) % K] += dp[prev][i];
      res[(i + 1) % K] += dp2[prev][i];
      res[(i + 2) % K] -= dp[curr][i];
    }
    res[K] += dp[prev][K] + dp2[prev][K] - dp[curr][K];
    res[K] += dp[prev][0] + dp2[prev][0] - dp[curr][0] - dp[curr][K - 1];
  }
  for (const auto& it : edge[curr])
    if (it != prev) {
      dfs2(it, curr);
    }
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = (int)(0); i < (int)(N - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs(1, -1);
  dfs2(1, -1);
  long long ans = 0;
  for (int i = (int)(1); i < (int)(N + 1); i++) {
    ans += dp[i][K] + dp2[i][K];
  }
  ans /= 2;
  printf("%lld\n", ans);
  return 0;
}

```