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;
const int maxn = 2e5;
const int INF = 1e9;
int head[maxn + 5], sz[maxn + 5], dis[maxn + 5], cnt[maxn + 5], tp[5];
bool vis[maxn + 5];
long long dp1[maxn + 5], dp2[maxn + 5], ans[5];
int root, mn, tot;
int n, k, u, v;
long long sum;
int num = 0;
struct edge {
  int to, _next;
} G[2 * maxn + 5];
void add(int u, int v) {
  G[++num].to = v;
  G[num]._next = head[u];
  head[u] = num;
}
void findrt(int u, int fa) {
  int num = 0;
  sz[u] = 1;
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (v == fa || vis[v]) continue;
    findrt(v, u);
    sz[u] += sz[v];
    if (num < sz[v]) num = sz[v];
  }
  if (tot - sz[u] > num) num = tot - sz[u];
  if (num < mn) {
    mn = num;
    root = u;
  }
}
void getDis(int u, int fa) {
  dis[u] = dis[fa] + 1;
  for (int i = 0; i < k; i++) ans[(dis[u] + i) % k] += tp[i];
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (v == fa || vis[v]) continue;
    getDis(v, u);
  }
}
void update(int u, int fa, bool op) {
  if (op)
    tp[dis[u] % k]++;
  else
    tp[dis[u] % k] = 0;
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (v == fa || vis[v]) continue;
    update(v, u, op);
  }
}
void calc(int u) {
  dis[u] = 0;
  tp[0] = 1;
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (vis[v]) continue;
    getDis(v, u);
    update(v, u, 1);
  }
  tp[0] = 0;
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (vis[v]) continue;
    update(v, u, 0);
  }
}
void divide(int u) {
  vis[u] = true;
  calc(u);
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (vis[v]) continue;
    tot = sz[v];
    mn = INF;
    findrt(v, 0);
    divide(root);
  }
}
void dfs1(int u, int fa) {
  cnt[u] = 1;
  dp1[u] = 0;
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (v == fa) continue;
    dfs1(v, u);
    cnt[u] += cnt[v];
    dp1[u] += dp1[v] + cnt[v];
  }
}
void dfs2(int u, int fa) {
  for (int i = head[u]; i; i = G[i]._next) {
    int v = G[i].to;
    if (v == fa) continue;
    dp2[v] = dp2[u] + (n - cnt[u]);
    dp2[v] = dp2[v] + dp1[u] - (dp1[v] + cnt[v]) + (cnt[u] - cnt[v]);
    dfs2(v, u);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  tot = n;
  mn = INF;
  findrt(1, 0);
  divide(root);
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= n; i++) sum = sum + dp1[i] + dp2[i];
  sum /= 2;
  for (int i = 1; i < k; i++) sum = sum + ans[i] * (k - i);
  printf("%I64d\n", sum / k);
  return 0;
}
