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.
#include <bits/stdc++.h>
using namespace std;
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
int n, k;
vector<int> g[200100];
vector<pair<int, int> > es;
int depth[200100];
long long chnum[200100];
long long dp[200100][5];
long long ans = 0ll;
void dfs(int v, int p, int d) {
  depth[v] = d;
  chnum[v] = 1ll;
  for (int i = 0; i < g[v].size(); i++) {
    if (g[v][i] == p) continue;
    dfs(g[v][i], v, d + 1);
    chnum[v] += chnum[g[v][i]];
  }
}
void rec(int v, int p) {
  for (int i = 0; i < g[v].size(); i++) {
    if (g[v][i] == p) continue;
    rec(g[v][i], v);
    dp[g[v][i]][0]++;
    for (int j = 0; j < k; j++) {
      for (int l = 0; l < k; l++) {
        ans += (dp[v][j] * dp[g[v][i]][l]) *
               (long long)((k * 2 - (j + l + 1)) % k);
      }
    }
    for (int j = 0; j < k; j++) {
      dp[v][(j + 1) % k] += dp[g[v][i]][j];
    }
  }
  for (int i = 1; i < k; i++) ans += dp[v][i] * (long long)(k - i);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
    es.push_back(pair<int, int>(a, b));
  }
  dfs(0, -1, 0);
  for (int i = 0; i < es.size(); i++) {
    int u = es[i].first, v = es[i].second;
    if (depth[u] > depth[v]) swap(u, v);
    ans += chnum[v] * (((long long)n) - chnum[v]);
  }
  rec(0, -1);
  assert(ans % (long long)k == 0ll);
  cout << ans / (long long)k << endl;
  return 0;
}
