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>
#pragma GCC optimize("Ofast")
using namespace std;
const int nmax = 200010;
vector<int> gr[nmax];
vector<int> parent(nmax, -1e9);
int dp[nmax][6];
int n, k;
int sz[nmax];
long long ans = 0;
int rem(int x, int y) { return (((x - y) % k) + k) % k; }
void dfs(int v, int pr, int mod, int depth) {
  parent[v] = pr;
  dp[v][mod] = 1;
  for (auto &to : gr[v]) {
    if (to == pr) continue;
    dfs(to, v, (mod + 1) % k, depth + 1);
    sz[v] += sz[to];
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= k; j++) {
        int ost1 = rem(i + j - 2, 2 * depth);
        int ost2 = rem(k, ost1);
        ans += 1LL * dp[to][j - 1] * ost2 * dp[v][i - 1];
      }
      dp[v][i - 1] += dp[to][i - 1];
    }
  }
}
int32_t main() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    --x;
    --y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= 5; j++) dp[i][j] = 0;
    sz[i] = 1;
  }
  dfs(0, -1e9, 0, 0);
  for (int i = 0; i < n; i++) ans += 1LL * sz[i] * (n - sz[i]);
  ans /= k;
  cout << ans;
}
