#include <bits/stdc++.h>
void sc(int &x) { scanf("%d", &x); }
void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
void sc(long long &x) { scanf("%lld", &x); }
void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void sc(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void sc(char *x) { scanf("%s", x); }
void sc(char *x, char *y) { scanf("%s%s", x, y); }
void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
void out(int x) { printf("%d\n", x); }
void out(long long x) { printf("%lld\n", x); }
void out(int x, int y) { printf("%d %d\n", x, y); }
void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
void out(long long x, long long y, long long z) {
  printf("%lld %lld %lld\n", x, y, z);
}
void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
using namespace std;
const int N = 205;
mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
int rdint(int l, int r) { return uniform_int_distribution<int>(l, r)(gen); }
int n, k, x[N], y[N];
pair<int, long long> dp[N][N];
int cal(long long m) {
  for (int len = 1; len < n; len++)
    for (int l = 0; l + len < n; l++) {
      int r = l + len;
      if (r == l + 1)
        dp[l][r] = {0, 0};
      else {
        dp[l][r] = {0, 0};
        for (int k = l + 1; k < r; k++) {
          long long s = abs(1ll * (y[r] - y[l]) * (x[k] - x[l]) -
                            1ll * (x[r] - x[l]) * (y[k] - y[l]));
          pair<int, long long> cur = {dp[l][k].first + dp[k][r].first,
                                      dp[l][k].second + dp[k][r].second + s};
          if (cur.second >= m) cur.first++, cur.second = 0;
          dp[l][r] = max(dp[l][r], cur);
        }
      }
    }
  return dp[0][n - 1].first;
}
void sol(int cas) {
  sc(n, k);
  for (int i = 0; i <= n - 1; i++) sc(x[i], y[i]);
  long long l = 1, r = 1e18, ans;
  while (l <= r) {
    long long m = (l + r) >> 1;
    if (cal(m) >= k + 1)
      ans = m, l = m + 1;
    else
      r = m - 1;
  }
  out(ans);
}
int main() {
  srand(time(0));
  int t = 1, cas = 0;
  while (t--) {
    sol(++cas);
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

You are given a strictly convex polygon with n vertices.

You will make k cuts that meet the following conditions: 

  * each cut is a segment that connects two different nonadjacent vertices; 
  * two cuts can intersect only at vertices of the polygon. 



Your task is to maximize the area of the smallest region that will be formed by the polygon and those k cuts.

Input

The first line contains two integers n, k (3 ≤ n ≤ 200, 0 ≤ k ≤ n-3).

The following n lines describe vertices of the polygon in anticlockwise direction. The i-th line contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^8) — the coordinates of the i-th vertex.

It is guaranteed that the polygon is convex and that no two adjacent sides are parallel.

Output

Print one integer: the maximum possible area of the smallest region after making k cuts multiplied by 2.

Examples

Input


8 4
-2 -4
2 -2
4 2
1 5
0 5
-4 4
-5 0
-5 -1


Output


11


Input


6 3
2 -2
2 -1
1 2
0 2
-2 1
-1 0


Output


3

Note

In the first example, it's optimal to make cuts between the following pairs of vertices: 

  * (-2, -4) and (4, 2), 
  * (-2, -4) and (1, 5), 
  * (-5, -1) and (1, 5), 
  * (-5, 0) and (0, 5). 

<image> Points (-5, -1), (1, 5), (0, 5), (-5, 0) determine the smallest region with double area of 11. 

In the second example, it's optimal to make cuts between the following pairs of vertices: 

  * (2, -1) and (0, 2), 
  * (2, -1) and (1, 0), 
  * (-1, 0) and (0, 2). 

<image> Points (2, -2), (2, -1), (-1, 0) determine one of the smallest regions with double area of 3. 

### Response

```cpp
#include <bits/stdc++.h>
void sc(int &x) { scanf("%d", &x); }
void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
void sc(long long &x) { scanf("%lld", &x); }
void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
void sc(long long &x, long long &y, long long &z) {
  scanf("%lld%lld%lld", &x, &y, &z);
}
void sc(char *x) { scanf("%s", x); }
void sc(char *x, char *y) { scanf("%s%s", x, y); }
void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
void out(int x) { printf("%d\n", x); }
void out(long long x) { printf("%lld\n", x); }
void out(int x, int y) { printf("%d %d\n", x, y); }
void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
void out(long long x, long long y, long long z) {
  printf("%lld %lld %lld\n", x, y, z);
}
void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
using namespace std;
const int N = 205;
mt19937 gen(chrono::system_clock::now().time_since_epoch().count());
int rdint(int l, int r) { return uniform_int_distribution<int>(l, r)(gen); }
int n, k, x[N], y[N];
pair<int, long long> dp[N][N];
int cal(long long m) {
  for (int len = 1; len < n; len++)
    for (int l = 0; l + len < n; l++) {
      int r = l + len;
      if (r == l + 1)
        dp[l][r] = {0, 0};
      else {
        dp[l][r] = {0, 0};
        for (int k = l + 1; k < r; k++) {
          long long s = abs(1ll * (y[r] - y[l]) * (x[k] - x[l]) -
                            1ll * (x[r] - x[l]) * (y[k] - y[l]));
          pair<int, long long> cur = {dp[l][k].first + dp[k][r].first,
                                      dp[l][k].second + dp[k][r].second + s};
          if (cur.second >= m) cur.first++, cur.second = 0;
          dp[l][r] = max(dp[l][r], cur);
        }
      }
    }
  return dp[0][n - 1].first;
}
void sol(int cas) {
  sc(n, k);
  for (int i = 0; i <= n - 1; i++) sc(x[i], y[i]);
  long long l = 1, r = 1e18, ans;
  while (l <= r) {
    long long m = (l + r) >> 1;
    if (cal(m) >= k + 1)
      ans = m, l = m + 1;
    else
      r = m - 1;
  }
  out(ans);
}
int main() {
  srand(time(0));
  int t = 1, cas = 0;
  while (t--) {
    sol(++cas);
  }
  return 0;
}

```