#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
struct rerooting {
  using F = function<T(T, int)>;
  using M = function<T(T, T)>;
  int V;
  vector<vector<int>> G;
  vector<vector<T>> dp;
  vector<T> ans;
  F f, g;
  M merge;
  T gen;
  rerooting(
      int V, F f, M merge, T gen, F g = [](T a, int b) { return a; })
      : V(V), f(f), merge(merge), gen(gen), g(g), G(V), dp(V), ans(V, gen) {}
  void add_edge(int a, int b) {
    G[a].push_back(b);
    G[b].push_back(a);
  }
  T dfs1(int pre, int now) {
    T ret = gen;
    for (int i = 0; i < G[now].size(); i++)
      if (G[now][i] != pre) {
        dp[now][i] = dfs1(now, G[now][i]);
        ret = merge(ret, f(dp[now][i], G[now][i]));
      }
    return g(ret, now);
  }
  void dfs2(int pre, int now, T frompar) {
    for (int i = 0; i < G[now].size(); i++) {
      if (G[now][i] == pre) {
        dp[now][i] = frompar;
        break;
      }
    }
    vector<T> lsum(G[now].size() + 1), rsum(G[now].size() + 1);
    lsum[0] = gen;
    rsum[G[now].size()] = gen;
    for (int i = 0; i < G[now].size(); i++)
      lsum[i + 1] = merge(lsum[i], f(dp[now][i], G[now][i]));
    for (int i = G[now].size(); i > 0; i--)
      rsum[i - 1] = merge(rsum[i], f(dp[now][i - 1], G[now][i - 1]));
    for (int i = 0; i < G[now].size(); i++)
      if (G[now][i] != pre) {
        T fromme = merge(lsum[i], rsum[i + 1]);
        dfs2(now, G[now][i], g(fromme, now));
      }
  }
  void build(int root = 0) {
    for (int i = 0; i < V; i++) dp[i].resize(G[i].size());
    dfs1(-1, root);
    dfs2(-1, root, gen);
    for (int i = 0; i < V; i++) {
      for (int j = 0; j < G[i].size(); j++)
        ans[i] = merge(ans[i], f(dp[i][j], G[i][j]));
      ans[i] = g(ans[i], i);
    }
  }
};
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  auto f = [&](const vector<ll> &a, int v) {
    vector<ll> ret(k + 1, 0);
    for (int i = 0; i < (k); i++) ret[i + 1] += a[i];
    ret[0]++;
    ret[k] += a[k];
    ret[0] += a[k - 1];
    return ret;
  };
  auto merge = [&](vector<ll> a, const vector<ll> &b) {
    for (int i = 0; i < (k + 1); i++) a[i] += b[i];
    return a;
  };
  vector<ll> gen(k + 1, 0);
  rerooting<vector<ll>> R(n, f, merge, gen);
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    R.add_edge(u, v);
  }
  R.build();
  ll ans = 0;
  for (int i = 0; i < (n); i++) {
    auto v = f(R.ans[i], 0);
    ll sum = 0;
    for (int j = 1; j <= k; j++) sum += v[j];
    ans += sum;
  }
  cout << ans / 2 << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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;
using ll = long long;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
struct rerooting {
  using F = function<T(T, int)>;
  using M = function<T(T, T)>;
  int V;
  vector<vector<int>> G;
  vector<vector<T>> dp;
  vector<T> ans;
  F f, g;
  M merge;
  T gen;
  rerooting(
      int V, F f, M merge, T gen, F g = [](T a, int b) { return a; })
      : V(V), f(f), merge(merge), gen(gen), g(g), G(V), dp(V), ans(V, gen) {}
  void add_edge(int a, int b) {
    G[a].push_back(b);
    G[b].push_back(a);
  }
  T dfs1(int pre, int now) {
    T ret = gen;
    for (int i = 0; i < G[now].size(); i++)
      if (G[now][i] != pre) {
        dp[now][i] = dfs1(now, G[now][i]);
        ret = merge(ret, f(dp[now][i], G[now][i]));
      }
    return g(ret, now);
  }
  void dfs2(int pre, int now, T frompar) {
    for (int i = 0; i < G[now].size(); i++) {
      if (G[now][i] == pre) {
        dp[now][i] = frompar;
        break;
      }
    }
    vector<T> lsum(G[now].size() + 1), rsum(G[now].size() + 1);
    lsum[0] = gen;
    rsum[G[now].size()] = gen;
    for (int i = 0; i < G[now].size(); i++)
      lsum[i + 1] = merge(lsum[i], f(dp[now][i], G[now][i]));
    for (int i = G[now].size(); i > 0; i--)
      rsum[i - 1] = merge(rsum[i], f(dp[now][i - 1], G[now][i - 1]));
    for (int i = 0; i < G[now].size(); i++)
      if (G[now][i] != pre) {
        T fromme = merge(lsum[i], rsum[i + 1]);
        dfs2(now, G[now][i], g(fromme, now));
      }
  }
  void build(int root = 0) {
    for (int i = 0; i < V; i++) dp[i].resize(G[i].size());
    dfs1(-1, root);
    dfs2(-1, root, gen);
    for (int i = 0; i < V; i++) {
      for (int j = 0; j < G[i].size(); j++)
        ans[i] = merge(ans[i], f(dp[i][j], G[i][j]));
      ans[i] = g(ans[i], i);
    }
  }
};
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  auto f = [&](const vector<ll> &a, int v) {
    vector<ll> ret(k + 1, 0);
    for (int i = 0; i < (k); i++) ret[i + 1] += a[i];
    ret[0]++;
    ret[k] += a[k];
    ret[0] += a[k - 1];
    return ret;
  };
  auto merge = [&](vector<ll> a, const vector<ll> &b) {
    for (int i = 0; i < (k + 1); i++) a[i] += b[i];
    return a;
  };
  vector<ll> gen(k + 1, 0);
  rerooting<vector<ll>> R(n, f, merge, gen);
  for (int i = 0; i < (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    R.add_edge(u, v);
  }
  R.build();
  ll ans = 0;
  for (int i = 0; i < (n); i++) {
    auto v = f(R.ans[i], 0);
    ll sum = 0;
    for (int j = 1; j <= k; j++) sum += v[j];
    ans += sum;
  }
  cout << ans / 2 << endl;
  return 0;
}

```