#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long k = 0;
  long long f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    k = (k << 1) + (k << 3) + (ch ^ 48);
    ch = getchar();
  }
  return k * f;
}
struct node {
  long long to, next;
} e[500050];
long long dp[200010][10];
long long sz[500050];
long long cnt, head[500054], ans, n, k;
void add(long long u, long long v) {
  ++cnt;
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long fa, long long dep) {
  sz[u] = dp[u][dep % k] = 1;
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    if (v == fa) continue;
    dfs(v, u, dep + 1);
    for (long long x = 0; x < k; x++) {
      for (long long y = 0; y < k; y++) {
        long long dis = ((x + y) % k - (dep * 2) % k + k) % k;
        ans += ((k - dis) % k) * dp[u][x] * dp[v][y];
      }
    }
    for (long long x = 0; x < k; x++) dp[u][x] += dp[v][x];
    ans += sz[v] * (n - sz[v]);
    sz[u] += sz[v];
  }
}
signed main() {
  n = read();
  k = read();
  for (long long i = 1; i < n; i++) {
    long long x, y;
    x = read();
    y = read();
    add(x, y);
    add(y, x);
  }
  dfs(1, 0, 0);
  cout << ans / k << endl;
}

### 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;
inline long long read() {
  long long k = 0;
  long long f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    k = (k << 1) + (k << 3) + (ch ^ 48);
    ch = getchar();
  }
  return k * f;
}
struct node {
  long long to, next;
} e[500050];
long long dp[200010][10];
long long sz[500050];
long long cnt, head[500054], ans, n, k;
void add(long long u, long long v) {
  ++cnt;
  e[cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long fa, long long dep) {
  sz[u] = dp[u][dep % k] = 1;
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    if (v == fa) continue;
    dfs(v, u, dep + 1);
    for (long long x = 0; x < k; x++) {
      for (long long y = 0; y < k; y++) {
        long long dis = ((x + y) % k - (dep * 2) % k + k) % k;
        ans += ((k - dis) % k) * dp[u][x] * dp[v][y];
      }
    }
    for (long long x = 0; x < k; x++) dp[u][x] += dp[v][x];
    ans += sz[v] * (n - sz[v]);
    sz[u] += sz[v];
  }
}
signed main() {
  n = read();
  k = read();
  for (long long i = 1; i < n; i++) {
    long long x, y;
    x = read();
    y = read();
    add(x, y);
    add(y, x);
  }
  dfs(1, 0, 0);
  cout << ans / k << endl;
}

```