#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int inf = (int)1e9 + 5;
const long long INF = (long long)1e18 + 5;
long long n, k;
vector<vector<long long>> a, out;
vector<bool> vis;
vector<long long> out_ans;
long long ans = 0;
long long dfs(long long v) {
  vis[v] = true;
  long long cur_ans = 0;
  for (auto to : a[v]) {
    if (vis[to]) continue;
    cur_ans += dfs(to);
    for (long long i = 0; i < k; i++) out[v][(i + 1) % k] += out[to][i];
    out[v][1 % k]++;
  }
  cur_ans += out[v][1 % k];
  out_ans[v] = cur_ans;
  ans += cur_ans;
  return cur_ans;
}
void dfs2(long long v, vector<long long> cur, long long cur_ans) {
  vis[v] = true;
  ans += cur_ans;
  (void)0;
  (void)0;
  (void)0;
  vector<long long> nxt(k, 0), nxt_copy;
  for (long long i = 0; i < k; i++) {
    nxt[i] += cur[i];
    nxt[i] += out[v][i];
  }
  nxt_copy = nxt;
  for (auto to : a[v]) {
    if (vis[to]) continue;
    nxt = nxt_copy;
    for (long long i = 0; i < k; i++) nxt[(i + 1) % k] -= out[to][i];
    nxt[1 % k]--;
    nxt[0]++;
    rotate(nxt.begin(), nxt.begin() + k - 1, nxt.end());
    long long nxt_ans =
        cur_ans + nxt[1 % k] + out_ans[v] - out_ans[to] - out[to][0] - 1;
    dfs2(to, nxt, nxt_ans);
  }
}
void init() {
  a.resize(n);
  vis.assign(n, false);
  out.assign(n, vector<long long>(k, 0));
  out_ans.resize(n);
}
void solve() {
  cin >> n >> k;
  init();
  long long x, y;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    x--;
    y--;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(0);
  fill(vis.begin(), vis.end(), false);
  dfs2(0, vector<long long>(k, 0), 0);
  cout << ans / 2;
}
int main() {
  string filename = "acm";
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(15);
  solve();
  (void)0;
  return 0;
}

### Prompt

Create a solution in cpp for 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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const int inf = (int)1e9 + 5;
const long long INF = (long long)1e18 + 5;
long long n, k;
vector<vector<long long>> a, out;
vector<bool> vis;
vector<long long> out_ans;
long long ans = 0;
long long dfs(long long v) {
  vis[v] = true;
  long long cur_ans = 0;
  for (auto to : a[v]) {
    if (vis[to]) continue;
    cur_ans += dfs(to);
    for (long long i = 0; i < k; i++) out[v][(i + 1) % k] += out[to][i];
    out[v][1 % k]++;
  }
  cur_ans += out[v][1 % k];
  out_ans[v] = cur_ans;
  ans += cur_ans;
  return cur_ans;
}
void dfs2(long long v, vector<long long> cur, long long cur_ans) {
  vis[v] = true;
  ans += cur_ans;
  (void)0;
  (void)0;
  (void)0;
  vector<long long> nxt(k, 0), nxt_copy;
  for (long long i = 0; i < k; i++) {
    nxt[i] += cur[i];
    nxt[i] += out[v][i];
  }
  nxt_copy = nxt;
  for (auto to : a[v]) {
    if (vis[to]) continue;
    nxt = nxt_copy;
    for (long long i = 0; i < k; i++) nxt[(i + 1) % k] -= out[to][i];
    nxt[1 % k]--;
    nxt[0]++;
    rotate(nxt.begin(), nxt.begin() + k - 1, nxt.end());
    long long nxt_ans =
        cur_ans + nxt[1 % k] + out_ans[v] - out_ans[to] - out[to][0] - 1;
    dfs2(to, nxt, nxt_ans);
  }
}
void init() {
  a.resize(n);
  vis.assign(n, false);
  out.assign(n, vector<long long>(k, 0));
  out_ans.resize(n);
}
void solve() {
  cin >> n >> k;
  init();
  long long x, y;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    x--;
    y--;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  dfs(0);
  fill(vis.begin(), vis.end(), false);
  dfs2(0, vector<long long>(k, 0), 0);
  cout << ans / 2;
}
int main() {
  string filename = "acm";
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(15);
  solve();
  (void)0;
  return 0;
}

```