#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 9;
int n, k, head[N], tot;
struct pp {
  int nxt, to;
} g[N << 1];
void add(int u, int v) { g[++tot].nxt = head[u], g[tot].to = v, head[u] = tot; }
int f[N][9], p[N], dep[N];
long long ans, sum;
int nxt(int x) { return (x - 1 < 0) ? (k - 1) : (x - 1); }
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  f[u][0]++;
  if (u != 1) ans += (dep[u] - 1) / k + 1;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    dfs(v, u);
    for (int j = 0; j < k; j++) f[u][j] += f[v][nxt(j)];
  }
  return;
}
void dfs2(int u, int fa, long long cnt) {
  sum += cnt;
  int t[9];
  for (int i = 0; i < k; i++) t[i] = 0;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    long long tpy = cnt;
    for (int j = 0; j < k; j++) t[j] = f[u][nxt(j)] - f[v][nxt(nxt(j))];
    tpy += t[1] - f[v][0];
    for (int j = 0; j < k; j++) f[v][j] += t[j];
    dfs2(v, u, tpy);
  }
  return;
}
int siz[N];
void dfs3(int u, int fa) {
  dep[u] = dep[fa] + 1, siz[u] = 1, ans += dep[u];
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    dfs3(v, u);
    siz[u] += siz[v];
  }
  return;
}
void dfs4(int u, int fa, long long cnt) {
  sum += cnt;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    dfs4(v, u, cnt + n - 2 * siz[v]);
  }
  return;
}
int main() {
  memset(head, -1, sizeof(head));
  tot = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  if (k == 1) {
    dep[0] = -1, ans = 0, sum = 0;
    dfs3(1, 0);
    dfs4(1, 0, ans);
    printf("%lld\n", sum / 2ll);
    return 0;
  }
  dep[0] = -1, ans = 0;
  dfs(1, 0);
  sum = 0;
  dfs2(1, 0, ans);
  printf("%lld\n", sum / 2ll);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 2e5 + 9;
int n, k, head[N], tot;
struct pp {
  int nxt, to;
} g[N << 1];
void add(int u, int v) { g[++tot].nxt = head[u], g[tot].to = v, head[u] = tot; }
int f[N][9], p[N], dep[N];
long long ans, sum;
int nxt(int x) { return (x - 1 < 0) ? (k - 1) : (x - 1); }
void dfs(int u, int fa) {
  dep[u] = dep[fa] + 1;
  f[u][0]++;
  if (u != 1) ans += (dep[u] - 1) / k + 1;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    dfs(v, u);
    for (int j = 0; j < k; j++) f[u][j] += f[v][nxt(j)];
  }
  return;
}
void dfs2(int u, int fa, long long cnt) {
  sum += cnt;
  int t[9];
  for (int i = 0; i < k; i++) t[i] = 0;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    long long tpy = cnt;
    for (int j = 0; j < k; j++) t[j] = f[u][nxt(j)] - f[v][nxt(nxt(j))];
    tpy += t[1] - f[v][0];
    for (int j = 0; j < k; j++) f[v][j] += t[j];
    dfs2(v, u, tpy);
  }
  return;
}
int siz[N];
void dfs3(int u, int fa) {
  dep[u] = dep[fa] + 1, siz[u] = 1, ans += dep[u];
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    dfs3(v, u);
    siz[u] += siz[v];
  }
  return;
}
void dfs4(int u, int fa, long long cnt) {
  sum += cnt;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    dfs4(v, u, cnt + n - 2 * siz[v]);
  }
  return;
}
int main() {
  memset(head, -1, sizeof(head));
  tot = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  if (k == 1) {
    dep[0] = -1, ans = 0, sum = 0;
    dfs3(1, 0);
    dfs4(1, 0, ans);
    printf("%lld\n", sum / 2ll);
    return 0;
  }
  dep[0] = -1, ans = 0;
  dfs(1, 0);
  sum = 0;
  dfs2(1, 0, ans);
  printf("%lld\n", sum / 2ll);
  return 0;
}

```