#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
struct Edge {
  int vet, nxt;
} e[N << 1];
int n, k, edge = 0;
int siz[N], head[N];
long long f[N][5];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void addedge(int x, int y) {
  edge++;
  e[edge].vet = y;
  e[edge].nxt = head[x];
  head[x] = edge;
}
inline void dfs1(int x, int fa) {
  siz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].vet;
    if (v == fa) continue;
    dfs1(v, x);
    siz[x] += siz[v];
    f[x][0] += f[v][k - 1] + siz[v];
    for (int j = 1; j <= k - 1; j++) f[x][j] += f[v][j - 1];
  }
}
inline void dfs2(int x, int fa) {
  long long t[5] = {};
  if (fa) {
    t[0] = f[fa][0] - f[x][k - 1] - siz[x];
    for (int i = 1; i < k; i++) t[i] = f[fa][i] - f[x][i - 1];
  }
  f[x][0] += t[k - 1] + n - siz[x];
  for (int i = 1; i < k; i++) f[x][i] += t[i - 1];
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].vet;
    if (v == fa) continue;
    dfs2(v, x);
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    addedge(x, y);
    addedge(y, x);
  }
  dfs1(1, 0), dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += f[i][0];
  printf("%lld\n", ans / 2);
  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 N = 200005;
struct Edge {
  int vet, nxt;
} e[N << 1];
int n, k, edge = 0;
int siz[N], head[N];
long long f[N][5];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void addedge(int x, int y) {
  edge++;
  e[edge].vet = y;
  e[edge].nxt = head[x];
  head[x] = edge;
}
inline void dfs1(int x, int fa) {
  siz[x] = 1;
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].vet;
    if (v == fa) continue;
    dfs1(v, x);
    siz[x] += siz[v];
    f[x][0] += f[v][k - 1] + siz[v];
    for (int j = 1; j <= k - 1; j++) f[x][j] += f[v][j - 1];
  }
}
inline void dfs2(int x, int fa) {
  long long t[5] = {};
  if (fa) {
    t[0] = f[fa][0] - f[x][k - 1] - siz[x];
    for (int i = 1; i < k; i++) t[i] = f[fa][i] - f[x][i - 1];
  }
  f[x][0] += t[k - 1] + n - siz[x];
  for (int i = 1; i < k; i++) f[x][i] += t[i - 1];
  for (int i = head[x]; i; i = e[i].nxt) {
    int v = e[i].vet;
    if (v == fa) continue;
    dfs2(v, x);
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    addedge(x, y);
    addedge(y, x);
  }
  dfs1(1, 0), dfs2(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += f[i][0];
  printf("%lld\n", ans / 2);
  return 0;
}

```