#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
struct Edge {
  int to, nex;
} edges[maxn << 1];
int head[maxn], cnt = 0, n, k;
inline void addEdge(int u, int v) {
  edges[++cnt].to = v;
  edges[cnt].nex = head[u];
  head[u] = cnt;
}
long long siz[maxn];
long long f[maxn][5], ans = 0ll;
void dfs(int u, int fa, int dep) {
  int v, rem, ad;
  siz[u] = 1;
  f[u][dep % k] = 1;
  for (int i = head[u]; i != -1; i = edges[i].nex) {
    v = edges[i].to;
    if (v == fa) continue;
    dfs(v, u, dep + 1);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        rem = ((i + j - 2 * dep) % k + k) % k;
        ad = ((k - rem) % k + k) % k;
        ans += (long long)ad * f[u][i] * f[v][j];
      }
    }
    for (int i = 0; i < k; i++) f[u][i] += f[v][i];
    siz[u] += siz[v];
  }
  ans += siz[u] * (n - siz[u]);
}
inline int read() {
  int x = 0;
  bool w = false;
  char c = 0;
  while (c > '9' || c < '0') {
    w |= c == '-';
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return w ? -x : x;
}
int main() {
  n = read();
  k = read();
  int u, v;
  memset(head, -1, sizeof(head));
  for (register int i = 1; i < n; i++) {
    u = read(), v = read();
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, 0, 0);
  printf("%lld", ans / (long long)k);
  return 0;
}

### Prompt

Generate 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 maxn = 200010;
struct Edge {
  int to, nex;
} edges[maxn << 1];
int head[maxn], cnt = 0, n, k;
inline void addEdge(int u, int v) {
  edges[++cnt].to = v;
  edges[cnt].nex = head[u];
  head[u] = cnt;
}
long long siz[maxn];
long long f[maxn][5], ans = 0ll;
void dfs(int u, int fa, int dep) {
  int v, rem, ad;
  siz[u] = 1;
  f[u][dep % k] = 1;
  for (int i = head[u]; i != -1; i = edges[i].nex) {
    v = edges[i].to;
    if (v == fa) continue;
    dfs(v, u, dep + 1);
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        rem = ((i + j - 2 * dep) % k + k) % k;
        ad = ((k - rem) % k + k) % k;
        ans += (long long)ad * f[u][i] * f[v][j];
      }
    }
    for (int i = 0; i < k; i++) f[u][i] += f[v][i];
    siz[u] += siz[v];
  }
  ans += siz[u] * (n - siz[u]);
}
inline int read() {
  int x = 0;
  bool w = false;
  char c = 0;
  while (c > '9' || c < '0') {
    w |= c == '-';
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return w ? -x : x;
}
int main() {
  n = read();
  k = read();
  int u, v;
  memset(head, -1, sizeof(head));
  for (register int i = 1; i < n; i++) {
    u = read(), v = read();
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, 0, 0);
  printf("%lld", ans / (long long)k);
  return 0;
}

```