#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f == -1 ? ~x + 1 : x;
}
long long n, k;
struct node {
  long long x, y;
} a[210];
pair<long long, long long> dp[210][210];
node operator-(node x, node y) { return node{x.x - y.x, x.y - y.y}; }
long long operator*(node x, node y) { return x.x * y.y - x.y * y.x; }
long long get(long long i, long long j, long long k) {
  return (a[j] - a[i]) * (a[k] - a[i]);
}
long long check(long long mid) {
  for (long long i = 1; i <= n; ++i) {
    for (long long j = i + 2; j <= n; ++j) dp[i][j] = make_pair(-1e8, -1e8);
  }
  for (long long l = 3; l <= n; ++l) {
    for (long long i = 1; i + l - 1 <= n; ++i) {
      long long j = i + l - 1;
      for (long long k = i; k <= j; ++k) {
        pair<long long, long long> nw =
            make_pair(dp[i][k].first + dp[k][j].first,
                      dp[i][k].second + dp[k][j].second + get(i, k, j));
        if (nw.second >= mid) ++nw.first, nw.second = 0;
        dp[i][j] = max(dp[i][j], nw);
      }
    }
  }
  return dp[1][n].first >= k + 1;
}
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; ++i) a[i].x = read(), a[i].y = read();
  long long l = 0, r = 1e18, ans = 0;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (check(mid))
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%lld\n", ans);
}

### Prompt

Your challenge is to write 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>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f == -1 ? ~x + 1 : x;
}
long long n, k;
struct node {
  long long x, y;
} a[210];
pair<long long, long long> dp[210][210];
node operator-(node x, node y) { return node{x.x - y.x, x.y - y.y}; }
long long operator*(node x, node y) { return x.x * y.y - x.y * y.x; }
long long get(long long i, long long j, long long k) {
  return (a[j] - a[i]) * (a[k] - a[i]);
}
long long check(long long mid) {
  for (long long i = 1; i <= n; ++i) {
    for (long long j = i + 2; j <= n; ++j) dp[i][j] = make_pair(-1e8, -1e8);
  }
  for (long long l = 3; l <= n; ++l) {
    for (long long i = 1; i + l - 1 <= n; ++i) {
      long long j = i + l - 1;
      for (long long k = i; k <= j; ++k) {
        pair<long long, long long> nw =
            make_pair(dp[i][k].first + dp[k][j].first,
                      dp[i][k].second + dp[k][j].second + get(i, k, j));
        if (nw.second >= mid) ++nw.first, nw.second = 0;
        dp[i][j] = max(dp[i][j], nw);
      }
    }
  }
  return dp[1][n].first >= k + 1;
}
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; ++i) a[i].x = read(), a[i].y = read();
  long long l = 0, r = 1e18, ans = 0;
  while (l <= r) {
    long long mid = l + r >> 1;
    if (check(mid))
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%lld\n", ans);
}

```