#include <bits/stdc++.h>
using namespace std;
template <typename T>
T getint() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * p;
}
mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
template <typename T1, typename T2>
bool umin(T1 &x, const T2 &y) {
  if (x > y) return x = y, true;
  return false;
}
template <typename T1, typename T2>
bool umax(T1 &x, const T2 &y) {
  if (x < y) return x = y, true;
  return false;
}
const int maxn = (int)3e5 + 10;
const int inf = (int)1e9 + 5;
const int mod = (int)1e9 + 7;
const long long llinf = (long long)1e18 + 5;
const double pi = acos(-1.0);
vector<int> g[maxn];
int was[maxn];
vector<int> nodes;
int par[maxn];
int sz[maxn];
void calcsz(int v, int p = -1) {
  nodes.push_back(v);
  sz[v] = 1;
  par[v] = p;
  for (int x : g[v]) {
    if (x == p || was[x]) continue;
    calcsz(x, v);
    sz[v] += sz[x];
  }
}
long long cnt[5];
long long sum[5];
long long res = 0;
int n;
int k;
void add(int x) {
  cnt[x % k]++;
  sum[x % k] += x / k;
}
void dfs2(int v, int par, int dep = 1) {
  add(dep);
  for (int x : g[v]) {
    if (x == par || was[x]) continue;
    dfs2(x, v, dep + 1);
  }
}
void dfs3(int v, int par, int dep = 1) {
  for (int j = (0); j < (k); j++) {
    int z = dep % k + j;
    res += sum[j];
    res += cnt[j] * 1LL * ((z > k ? 2 : (z == 0 ? 0 : 1)) + (dep / k));
  }
  for (int x : g[v]) {
    if (x == par || was[x]) continue;
    dfs3(x, v, dep + 1);
  }
}
void build(int v, int depth = 1) {
  nodes.clear();
  calcsz(v);
  int c = -1;
  for (int x : nodes) {
    int mxsz = nodes.size() - sz[x];
    for (int v : g[x]) {
      if (v != par[x] && !was[v]) umax(mxsz, sz[v]);
    }
    if (mxsz <= nodes.size() / 2) c = x;
  }
  for (int j = (0); j < (5); j++) cnt[j] = 0, sum[j] = 0;
  add(0);
  for (int x : g[c]) {
    if (was[x]) continue;
    dfs3(x, c);
    dfs2(x, c);
  }
  was[c] = 1;
  for (int x : g[c]) {
    if (was[x]) continue;
    build(x, depth + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = (0); i < (n - 1); i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  build(1);
  cout << res << endl;
  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;
template <typename T>
T getint() {
  T x = 0, p = 1;
  char ch;
  do {
    ch = getchar();
  } while (ch <= ' ');
  if (ch == '-') p = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * p;
}
mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
template <typename T1, typename T2>
bool umin(T1 &x, const T2 &y) {
  if (x > y) return x = y, true;
  return false;
}
template <typename T1, typename T2>
bool umax(T1 &x, const T2 &y) {
  if (x < y) return x = y, true;
  return false;
}
const int maxn = (int)3e5 + 10;
const int inf = (int)1e9 + 5;
const int mod = (int)1e9 + 7;
const long long llinf = (long long)1e18 + 5;
const double pi = acos(-1.0);
vector<int> g[maxn];
int was[maxn];
vector<int> nodes;
int par[maxn];
int sz[maxn];
void calcsz(int v, int p = -1) {
  nodes.push_back(v);
  sz[v] = 1;
  par[v] = p;
  for (int x : g[v]) {
    if (x == p || was[x]) continue;
    calcsz(x, v);
    sz[v] += sz[x];
  }
}
long long cnt[5];
long long sum[5];
long long res = 0;
int n;
int k;
void add(int x) {
  cnt[x % k]++;
  sum[x % k] += x / k;
}
void dfs2(int v, int par, int dep = 1) {
  add(dep);
  for (int x : g[v]) {
    if (x == par || was[x]) continue;
    dfs2(x, v, dep + 1);
  }
}
void dfs3(int v, int par, int dep = 1) {
  for (int j = (0); j < (k); j++) {
    int z = dep % k + j;
    res += sum[j];
    res += cnt[j] * 1LL * ((z > k ? 2 : (z == 0 ? 0 : 1)) + (dep / k));
  }
  for (int x : g[v]) {
    if (x == par || was[x]) continue;
    dfs3(x, v, dep + 1);
  }
}
void build(int v, int depth = 1) {
  nodes.clear();
  calcsz(v);
  int c = -1;
  for (int x : nodes) {
    int mxsz = nodes.size() - sz[x];
    for (int v : g[x]) {
      if (v != par[x] && !was[v]) umax(mxsz, sz[v]);
    }
    if (mxsz <= nodes.size() / 2) c = x;
  }
  for (int j = (0); j < (5); j++) cnt[j] = 0, sum[j] = 0;
  add(0);
  for (int x : g[c]) {
    if (was[x]) continue;
    dfs3(x, c);
    dfs2(x, c);
  }
  was[c] = 1;
  for (int x : g[c]) {
    if (was[x]) continue;
    build(x, depth + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  for (int i = (0); i < (n - 1); i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  build(1);
  cout << res << endl;
  return 0;
}

```