#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n & 1) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n & 1) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
vector<int> prime;
bool primee[10000000];
void primegen(long long x) {
  for (long long i = 2; i * i <= x; i++)
    if (!primee[i])
      for (long long j = i * i; j <= x; j += i) primee[j] = true;
  for (long long i = 2; i <= x; i++)
    if (!primee[i]) prime.push_back(i);
}
const int N = 234567;
long long coun[6], dp[N][6], ans;
vector<int> v[N];
int sz[N], k, n;
void dfs(int s, int pa) {
  dp[s][0] = 1;
  sz[s] = 1;
  int i, j, l;
  for (i = 0; i < v[s].size(); i++)
    if (v[s][i] != pa) {
      int u = v[s][i];
      dfs(u, s);
      sz[s] += sz[u];
      for (j = 0; j < k; j++)
        for (l = 0; l < k; l++) coun[(j + l + 1) % k] += (dp[s][j] * dp[u][l]);
      for (j = 0; j < k; j++) dp[s][(j + 1) % k] += dp[u][j];
    }
  ans += (n - (long long)sz[s]) * (long long)sz[s];
}
int main() {
  int i, j;
  scanf("%d", &n);
  scanf("%d", &k);
  int x, y;
  for (i = 0; i < n - 1; i++) {
    scanf("%d", &x);
    scanf("%d", &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  for (i = 1; i < k; i++) ans += (k - i) * coun[i];
  printf("%lld\n", ans / k);
  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;
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n & 1) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n & 1) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
vector<int> prime;
bool primee[10000000];
void primegen(long long x) {
  for (long long i = 2; i * i <= x; i++)
    if (!primee[i])
      for (long long j = i * i; j <= x; j += i) primee[j] = true;
  for (long long i = 2; i <= x; i++)
    if (!primee[i]) prime.push_back(i);
}
const int N = 234567;
long long coun[6], dp[N][6], ans;
vector<int> v[N];
int sz[N], k, n;
void dfs(int s, int pa) {
  dp[s][0] = 1;
  sz[s] = 1;
  int i, j, l;
  for (i = 0; i < v[s].size(); i++)
    if (v[s][i] != pa) {
      int u = v[s][i];
      dfs(u, s);
      sz[s] += sz[u];
      for (j = 0; j < k; j++)
        for (l = 0; l < k; l++) coun[(j + l + 1) % k] += (dp[s][j] * dp[u][l]);
      for (j = 0; j < k; j++) dp[s][(j + 1) % k] += dp[u][j];
    }
  ans += (n - (long long)sz[s]) * (long long)sz[s];
}
int main() {
  int i, j;
  scanf("%d", &n);
  scanf("%d", &k);
  int x, y;
  for (i = 0; i < n - 1; i++) {
    scanf("%d", &x);
    scanf("%d", &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  for (i = 1; i < k; i++) ans += (k - i) * coun[i];
  printf("%lld\n", ans / k);
  return 0;
}

```