#include <bits/stdc++.h>
using namespace std;
struct node {
  int v, next;
} p[200005 << 1];
int head[200005], n, k;
long long dp[200005][10], size[200005], ans;
void dfs(int x, int fa, int dep) {
  size[x] = dp[x][dep % k] = 1;
  for (int i = head[x]; ~i; i = p[i].next) {
    if (p[i].v == fa) continue;
    dfs(p[i].v, x, dep + 1);
    for (int a = 0; a < k; a++) {
      for (int b = 0; b < k; b++) {
        int dis = ((a + b) % k - (dep * 2) % k + k) % k;
        ans += ((k - dis) % k) * dp[x][a] * dp[p[i].v][b];
      }
    }
    for (int a = 0; a < k; a++) {
      dp[x][a] += dp[p[i].v][a];
    }
    ans += size[p[i].v] * (n - size[p[i].v]);
    size[x] += size[p[i].v];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  memset(head, -1, sizeof(head));
  int len = 0;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    p[len].v = v;
    p[len].next = head[u];
    head[u] = len++;
    p[len].v = u;
    p[len].next = head[v];
    head[v] = len++;
  }
  dfs(1, 0, 0);
  printf("%lld\n", ans / k);
}

### 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;
struct node {
  int v, next;
} p[200005 << 1];
int head[200005], n, k;
long long dp[200005][10], size[200005], ans;
void dfs(int x, int fa, int dep) {
  size[x] = dp[x][dep % k] = 1;
  for (int i = head[x]; ~i; i = p[i].next) {
    if (p[i].v == fa) continue;
    dfs(p[i].v, x, dep + 1);
    for (int a = 0; a < k; a++) {
      for (int b = 0; b < k; b++) {
        int dis = ((a + b) % k - (dep * 2) % k + k) % k;
        ans += ((k - dis) % k) * dp[x][a] * dp[p[i].v][b];
      }
    }
    for (int a = 0; a < k; a++) {
      dp[x][a] += dp[p[i].v][a];
    }
    ans += size[p[i].v] * (n - size[p[i].v]);
    size[x] += size[p[i].v];
  }
}
int main() {
  scanf("%d%d", &n, &k);
  memset(head, -1, sizeof(head));
  int len = 0;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    p[len].v = v;
    p[len].next = head[u];
    head[u] = len++;
    p[len].v = u;
    p[len].next = head[v];
    head[v] = len++;
  }
  dfs(1, 0, 0);
  printf("%lld\n", ans / k);
}

```