#include <bits/stdc++.h>
using namespace std;
int Y[] = {-1, 1, 0, 0, -1, -1, 1, 1};
int X[] = {0, 0, -1, 1, -1, 1, -1, 1};
int directions = 4;
long long power(long long a, long long n) {
  if (a == 0) return 0;
  if (a == 1 || n == 0) return 1;
  if (n == 1) return a % 1000000007;
  long long t = power(a, n / 2);
  t = t * t % 1000000007;
  if (n & 1) return t * a % 1000000007;
  return t;
}
void keywords() {
  for (int i = 1; i < 2; i++) continue;
  for (int i = 1; i < 2; i++) break;
  return;
}
int popcount(long long a) {
  int c = 0;
  while (a) {
    c++;
    a -= a & -a;
  }
  return c;
}
void factorize(int a) {}
void update(int tree[], int idx, int val, int maxval) {
  for (; idx <= maxval; idx += idx & -idx) {
    tree[idx] += val;
  }
}
int read(int tree[], int idx) {
  long long sum = 0;
  for (; idx > 0; idx -= idx & -idx) {
    sum += tree[idx];
  }
  return sum;
}
struct node2 {
  int id, val;
  node2() {
    static int ctx = 1;
    id = ctx++;
  };
  node2(int a, int b = 0, int c = 0, int d = 0, int e = 0, int f = 0) {
    val = a;
  }
};
struct comp2 {
  bool operator()(int a, int b) { return b < a; }
};
bool cmp2(int a, int b) { return b < a; }
struct node {
  int id, val;
  node() {
    static int ctx = 1;
    id = ctx++;
  };
  node(int a, int b = 0, int c = 0, int d = 0, int e = 0, int f = 0) {
    val = a;
  }
};
struct comp {
  bool operator()(int a, int b) { return b < a; }
};
bool cmp(int a, int b) { return b < a; }
long long n, m, a, b, c, d, k, h, w, x, y, p, q, t, ans, res, ma, mi, T,
    act = 0, pas = 1, cur, flag, now;
long long jumps[200009][10], cnt[200009][10];
char s[1];
double e, f, z;
vector<int> v[200009], vec;
set<int> sett;
map<int, int> mapp;
void print() {}
void print2() {}
void input() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
}
void init(int id, int par) {
  for (auto u : v[id]) {
    if (u == par) continue;
    init(u, id);
    for (int rem = 1; rem < k; rem++) {
      jumps[id][(rem + 1) % k] += jumps[u][rem];
      cnt[id][(rem + 1) % k] += cnt[u][rem];
    }
    jumps[id][1 % k] += jumps[u][0] + 1 * cnt[u][0] + 1;
    cnt[id][1 % k] += cnt[u][0] + 1;
    cnt[u][0] += 1;
  }
}
void calc(int id, int par) {
  for (int rem = 0; rem < k; rem++) ans += jumps[id][rem];
  long long sum[10] = {0}, jgsfhksadvk[10] = {0};
  for (auto u : v[id]) {
    if (u == par) continue;
    calc(u, id);
    for (int remp = 0; remp < k; remp++) {
      for (int remn = 0; remn < k; remn++) {
        ans += sum[remp] * cnt[u][remn] + jumps[u][remn] * jgsfhksadvk[remp];
        if (remp == 0 && remn == 0)
          ans += (k == 1 ? 2 : 1) * cnt[u][0] * jgsfhksadvk[0];
        else if (remp == 0 || remn == 0) {
          if (remp + remn + 2 > k) ans += cnt[u][remn] * jgsfhksadvk[remp];
        } else {
          if (remp + remn + 2 <= k) ans -= cnt[u][remn] * jgsfhksadvk[remp];
        }
      }
    }
    for (int rem = 0; rem < k; rem++) {
      sum[rem] += jumps[u][rem];
      jgsfhksadvk[rem] += cnt[u][rem];
    }
  }
}
void solve() {
  init(1, 0);
  calc(1, 0);
}
void output() { cout << ans; }
int main() {
  input();
  solve();
  output();
  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;
int Y[] = {-1, 1, 0, 0, -1, -1, 1, 1};
int X[] = {0, 0, -1, 1, -1, 1, -1, 1};
int directions = 4;
long long power(long long a, long long n) {
  if (a == 0) return 0;
  if (a == 1 || n == 0) return 1;
  if (n == 1) return a % 1000000007;
  long long t = power(a, n / 2);
  t = t * t % 1000000007;
  if (n & 1) return t * a % 1000000007;
  return t;
}
void keywords() {
  for (int i = 1; i < 2; i++) continue;
  for (int i = 1; i < 2; i++) break;
  return;
}
int popcount(long long a) {
  int c = 0;
  while (a) {
    c++;
    a -= a & -a;
  }
  return c;
}
void factorize(int a) {}
void update(int tree[], int idx, int val, int maxval) {
  for (; idx <= maxval; idx += idx & -idx) {
    tree[idx] += val;
  }
}
int read(int tree[], int idx) {
  long long sum = 0;
  for (; idx > 0; idx -= idx & -idx) {
    sum += tree[idx];
  }
  return sum;
}
struct node2 {
  int id, val;
  node2() {
    static int ctx = 1;
    id = ctx++;
  };
  node2(int a, int b = 0, int c = 0, int d = 0, int e = 0, int f = 0) {
    val = a;
  }
};
struct comp2 {
  bool operator()(int a, int b) { return b < a; }
};
bool cmp2(int a, int b) { return b < a; }
struct node {
  int id, val;
  node() {
    static int ctx = 1;
    id = ctx++;
  };
  node(int a, int b = 0, int c = 0, int d = 0, int e = 0, int f = 0) {
    val = a;
  }
};
struct comp {
  bool operator()(int a, int b) { return b < a; }
};
bool cmp(int a, int b) { return b < a; }
long long n, m, a, b, c, d, k, h, w, x, y, p, q, t, ans, res, ma, mi, T,
    act = 0, pas = 1, cur, flag, now;
long long jumps[200009][10], cnt[200009][10];
char s[1];
double e, f, z;
vector<int> v[200009], vec;
set<int> sett;
map<int, int> mapp;
void print() {}
void print2() {}
void input() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
}
void init(int id, int par) {
  for (auto u : v[id]) {
    if (u == par) continue;
    init(u, id);
    for (int rem = 1; rem < k; rem++) {
      jumps[id][(rem + 1) % k] += jumps[u][rem];
      cnt[id][(rem + 1) % k] += cnt[u][rem];
    }
    jumps[id][1 % k] += jumps[u][0] + 1 * cnt[u][0] + 1;
    cnt[id][1 % k] += cnt[u][0] + 1;
    cnt[u][0] += 1;
  }
}
void calc(int id, int par) {
  for (int rem = 0; rem < k; rem++) ans += jumps[id][rem];
  long long sum[10] = {0}, jgsfhksadvk[10] = {0};
  for (auto u : v[id]) {
    if (u == par) continue;
    calc(u, id);
    for (int remp = 0; remp < k; remp++) {
      for (int remn = 0; remn < k; remn++) {
        ans += sum[remp] * cnt[u][remn] + jumps[u][remn] * jgsfhksadvk[remp];
        if (remp == 0 && remn == 0)
          ans += (k == 1 ? 2 : 1) * cnt[u][0] * jgsfhksadvk[0];
        else if (remp == 0 || remn == 0) {
          if (remp + remn + 2 > k) ans += cnt[u][remn] * jgsfhksadvk[remp];
        } else {
          if (remp + remn + 2 <= k) ans -= cnt[u][remn] * jgsfhksadvk[remp];
        }
      }
    }
    for (int rem = 0; rem < k; rem++) {
      sum[rem] += jumps[u][rem];
      jgsfhksadvk[rem] += cnt[u][rem];
    }
  }
}
void solve() {
  init(1, 0);
  calc(1, 0);
}
void output() { cout << ans; }
int main() {
  input();
  solve();
  output();
  return 0;
}

```