You are given two integers n and k.

You need to construct k regular polygons having same [circumcircle](https://en.wikipedia.org/wiki/Circumscribed_circle), with distinct number of sides l between 3 and n. 

<image> Illustration for the first example.

You can rotate them to minimize the total number of distinct points on the circle. Find the minimum number of such points.

Input

The only line of input contains two integers n and k (3 ≤ n ≤ 10^{6}, 1 ≤ k ≤ n-2), the maximum number of sides of a polygon and the number of polygons to construct, respectively.

Output

Print a single integer — the minimum number of points required for k polygons.

Examples

Input


6 2


Output


6


Input


200 50


Output


708

Note

In the first example, we have n = 6 and k = 2. So, we have 4 polygons with number of sides 3, 4, 5 and 6 to choose from and if we choose the triangle and the hexagon, then we can arrange them as shown in the picture in the statement.

Hence, the minimum number of points required on the circle is 6, which is also the minimum overall possible sets.
#include <bits/stdc++.h>
using namespace std;
namespace mine {
long long qread() {
  long long ans = 0, f = 1;
  char c = getchar();
  while (c < '0' or c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c and c <= '9') ans = ans * 10 + c - '0', c = getchar();
  return ans * f;
}
void write(long long num) {
  if (num < 0) putchar('-'), num = -num;
  if (num >= 10) write(num / 10);
  putchar('0' + num % 10);
}
void write1(long long num) {
  write(num);
  putchar(' ');
}
void write2(long long num) {
  write(num);
  putchar('\n');
}
void chmax(int &x, const long long y) { x = (x > y ? x : y); }
void chmin(int &x, const long long y) { x = (x < y ? x : y); }
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : 0; }
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 9;
inline int mm(const int x) { return x >= MOD ? x - MOD : x; }
inline long long qpower(long long x, long long e) {
  long long ans = 1;
  while (e) {
    if (e & 1) ans = ans * x % MOD;
    x = x * x % MOD;
    e >>= 1;
  }
  return ans;
}
inline int invm(int x) { return qpower(x, MOD - 2); }
const int N = 3e6 + 10;
bool nop[N];
int pp, prn[N];
int phi[N];
void pre() {
  phi[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!nop[i]) prn[++pp] = i, phi[i] = i - 1;
    for (int j = 1; j <= pp and (long long) i * prn[j] < N; j++) {
      int t = i * prn[j];
      nop[t] = 1;
      if (i % prn[j] == 0) {
        phi[t] = phi[i] * prn[j];
        break;
      }
      phi[t] = phi[i] * (prn[j] - 1);
    }
  }
}
void main() {
  pre();
  int n = qread(), k = qread() + 2;
  if (k == 3) {
    write(3);
    return;
  }
  sort(phi + 1, phi + n + 1);
  long long ans = 0;
  for (int i = 1; i <= k; i++) ans += phi[i];
  write(ans);
}
};  // namespace mine
int main() {
  srand(time(0));
  mine::main();
}
