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;
int64_t n, k;
const int64_t maxn = 2e5 + 42, maxk = 10;
vector<int64_t> g[maxn];
int64_t sz[maxn][maxk];
int64_t up[maxn][maxk];
int64_t ans = 0;
int64_t penalty = 0;
void dfs1(int64_t v = 0, int64_t p = 0) {
  sz[v][0] = 1;
  for (auto u : g[v]) {
    if (u == p) continue;
    dfs1(u, v);
    for (int64_t z = 0; z < k; z++) sz[v][(z + 1) % k] += sz[u][z];
  }
}
void dfs2(int64_t v = 0, int64_t p = 0) {
  for (auto u : g[v]) {
    if (u == p) continue;
    for (int64_t z = 0; z < k; z++)
      up[u][(z + 1) % k] = up[v][z] + sz[v][z] - sz[u][(z + k - 1) % k];
    dfs2(u, v);
  }
  if (v != p) {
    int64_t sm = accumulate(sz[v], sz[v] + k, 0);
    int64_t smu = accumulate(up[v], up[v] + k, 0);
    ans += smu * sz[v][0] + sm * up[v][1 % k];
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int64_t i = 0; i < n - 1; i++) {
    int64_t u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs1();
  dfs2();
  cout << ans / 2 << endl;
}
