#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200010;
struct Edge {
  int u, v, next;
} e[MAXN * 2];
int head[MAXN], tot;
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void add_edge(int u, int v) {
  e[tot].u = u, e[tot].v = v, e[tot].next = head[u];
  head[u] = tot++;
}
int n, k;
long long int dp[MAXN][6], size[MAXN];
long long int ans = 0, tmp = 0;
void dfs(int u, int fa) {
  size[u] = 1;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
    ans += size[v] * (n - size[v]);
    size[u] += size[v];
    for (int j = 0; j < k; j++) {
      for (int l = 0; l < k; l++) {
        long long int ta = dp[u][j];
        long long int tb = dp[v][l];
        if (l == 0) tb++;
        tmp += ((k - (j + l + 1) % k) % k) * ta * tb;
      }
    }
    for (int j = 0; j < k; j++) dp[u][j] += dp[v][(j - 1 + k) % k];
    dp[u][1]++;
  }
  for (int i = 1; i < k; i++) tmp += (k - i) * dp[u][i];
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  init();
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
  }
  memset(size, 0, sizeof(size));
  dfs(1, 1);
  if (k == 1) {
    printf("%I64d\n", ans);
    return 0;
  }
  printf("%I64d\n", (ans + tmp) / k);
  return 0;
}

### Prompt

Your task is to create 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 u, v, next;
} e[MAXN * 2];
int head[MAXN], tot;
void init() {
  memset(head, -1, sizeof(head));
  tot = 0;
}
void add_edge(int u, int v) {
  e[tot].u = u, e[tot].v = v, e[tot].next = head[u];
  head[u] = tot++;
}
int n, k;
long long int dp[MAXN][6], size[MAXN];
long long int ans = 0, tmp = 0;
void dfs(int u, int fa) {
  size[u] = 1;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
    ans += size[v] * (n - size[v]);
    size[u] += size[v];
    for (int j = 0; j < k; j++) {
      for (int l = 0; l < k; l++) {
        long long int ta = dp[u][j];
        long long int tb = dp[v][l];
        if (l == 0) tb++;
        tmp += ((k - (j + l + 1) % k) % k) * ta * tb;
      }
    }
    for (int j = 0; j < k; j++) dp[u][j] += dp[v][(j - 1 + k) % k];
    dp[u][1]++;
  }
  for (int i = 1; i < k; i++) tmp += (k - i) * dp[u][i];
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  init();
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
  }
  memset(size, 0, sizeof(size));
  dfs(1, 1);
  if (k == 1) {
    printf("%I64d\n", ans);
    return 0;
  }
  printf("%I64d\n", (ans + tmp) / k);
  return 0;
}

```