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;
inline void pi(int x) {
  if (x == 0) {
    putchar('0');
    return;
  }
  char c[10];
  int k = 0;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  while (x) {
    int nx = x / 10;
    c[k++] = x - nx * 10;
    x = nx;
  }
  while (k--) {
    putchar(c[k] + '0');
  }
}
inline void pin(const int& x) {
  pi(x);
  putchar('\n');
};
inline void pl(long long x) {
  if (x == 0) {
    putchar('0');
    return;
  }
  char c[19];
  int k = 0;
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  while (x) {
    long long nx = x / 10;
    c[k++] = x - nx * 10;
    x = nx;
  }
  while (k--) {
    putchar(c[k] + '0');
  }
}
inline void pln(const long long& x) {
  pl(x);
  putchar('\n');
};
inline void gi(int& val) {
  val = 1;
  char c;
  do {
    c = getchar();
  } while (c != '-' && (c < '0' || c > '9'));
  int zn = 1;
  if (c == '-') {
    zn = -1;
    c = getchar();
  }
  val = c - '0';
  c = getchar();
  while (c >= '0' && c <= '9') {
    val *= 10;
    val += c - '0';
    c = getchar();
  }
  val *= zn;
}
inline void gl(long long& val) {
  val = 1;
  char c;
  do {
    c = getchar();
  } while (c != '-' && (c < '0' || c > '9'));
  int zn = 1;
  if (c == '-') {
    zn = -1;
    c = getchar();
  }
  val = c - '0';
  c = getchar();
  while (c >= '0' && c <= '9') {
    val *= 10;
    val += c - '0';
    c = getchar();
  }
  val *= zn;
}
inline void gl(long long& val1, long long& val2) {
  gl(val1);
  gl(val2);
}
inline void gi(int& val1, int& val2) {
  gi(val1);
  gi(val2);
}
inline void gl(long long& val1, long long& val2, long long& val3) {
  gl(val1);
  gl(val2);
  gl(val3);
}
inline void gi(int& val1, int& val2, int& val3) {
  gi(val1);
  gi(val2);
  gi(val3);
}
inline void mygets(char* b) {
  int i = 0;
  for (;; ++i) {
    const char c = getchar();
    if (c == '\n' || c == EOF) {
      break;
    }
    b[i] = c;
  }
  b[i] = '\0';
}
inline void mygets(string& b) {
  int i = 0;
  for (;; ++i) {
    const char c = getchar();
    if (c == '\n' || c == EOF) {
      break;
    }
    b.push_back(c);
  }
}
inline void mygets(char& b) {
  b = getchar();
  getchar();
}
inline void myputs(const string& b) {
  for (int i = 0; i < b.size(); ++i) {
    putchar(b[i]);
  }
}
inline void putsn(const string& b) {
  myputs(b);
  putchar('\n');
}
inline void myputs(const char& b) { putchar(b); }
inline void putsn(const char& b) {
  putchar(b);
  putchar('\n');
}
const int N = 2e5 + 7, M = 150000, mod = 1000000007, inf = 1000000007;
const long long lmod = 1000000007, linf = 1e18 + 77;
const long double eps = 1e-14;
vector<int> g[N];
bool used[N];
int n, k;
long long r[N][5];
int c[N][5];
long long ans = 0;
void go(int v) {
  used[v] = 1;
  c[v][0] = 1;
  for (auto to : g[v]) {
    if (!used[to]) {
      go(to);
      for (int i = k - 1; i > 0; i--) {
        swap(r[to][i], r[to][i - 1]);
        swap(c[to][i], c[to][i - 1]);
      }
      for (int i = 0; i < k; i++) {
        r[to][i] += c[to][i];
      }
      for (int i = 0; i < k; i++) {
        for (int j = 0; j < k; j++) {
          int ost = (i + j) % k;
          if (ost == 0) {
            ans += (r[v][i] * c[to][j] + r[to][j] * c[v][i]) / k;
          } else {
            ans += (r[v][i] * c[to][j] + r[to][j] * c[v][i] -
                    (long long)ost * c[to][j] * c[v][i]) /
                       k +
                   (long long)c[to][j] * c[v][i];
          }
        }
      }
      for (int i = 0; i < k; i++) {
        r[v][i] += r[to][i];
        c[v][i] += c[to][i];
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  gi(n, k);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    gi(u, v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  go(1);
  pln(ans);
}
