#include <bits/stdc++.h>
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
using namespace std;
long long ksm(long long a, long long b) {
  if (!b) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 998244353;
  if (b & 1) ns = ns * a % 998244353;
  return ns;
}
pair<int, long long> dp[205][205];
long long x[205], y[205], S;
int nxt[205], n, k;
int check(long long s) {
  S = s;
  int d = 0;
  for (int dis = 1; dis < n; dis++) {
    for (int r = 0; r < n; r++) {
      int l = (r + dis) % n;
      if (dis == 1) {
        dp[l][r] = make_pair(0, 0);
        continue;
      }
      dp[l][r] = make_pair(-1, -1);
      for (int k = nxt[r]; k != l; k = nxt[k]) {
        long long s =
            (x[r] - x[l]) * (y[k] - y[l]) - (x[k] - x[l]) * (y[r] - y[l]);
        auto s1 = dp[l][k], s2 = dp[k][r];
        int p1 = s1.first + s2.first;
        long long p2 = s1.second + s2.second + s;
        if (p2 >= S) p2 = 0, p1++;
        dp[l][r] = max(dp[l][r], make_pair(p1, p2));
      }
    }
  }
  for (int i = 0; i < n; i++) {
    d = max(d, dp[i][nxt[i]].first);
  }
  return d;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++)
    scanf("%lld%lld", x + i, y + i), nxt[i] = (i + 1) % n;
  long long l = 0, r = (1ll << 60);
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (check(mid) >= k + 1)
      l = mid;
    else
      r = mid;
  }
  printf("%lld\n", l);
}

### Prompt

In cpp, your task is to solve 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>
template <typename T>
bool chkmax(T &x, T y) {
  return x < y ? x = y, true : false;
}
template <typename T>
bool chkmin(T &x, T y) {
  return x > y ? x = y, true : false;
}
using namespace std;
long long ksm(long long a, long long b) {
  if (!b) return 1;
  long long ns = ksm(a, b >> 1);
  ns = ns * ns % 998244353;
  if (b & 1) ns = ns * a % 998244353;
  return ns;
}
pair<int, long long> dp[205][205];
long long x[205], y[205], S;
int nxt[205], n, k;
int check(long long s) {
  S = s;
  int d = 0;
  for (int dis = 1; dis < n; dis++) {
    for (int r = 0; r < n; r++) {
      int l = (r + dis) % n;
      if (dis == 1) {
        dp[l][r] = make_pair(0, 0);
        continue;
      }
      dp[l][r] = make_pair(-1, -1);
      for (int k = nxt[r]; k != l; k = nxt[k]) {
        long long s =
            (x[r] - x[l]) * (y[k] - y[l]) - (x[k] - x[l]) * (y[r] - y[l]);
        auto s1 = dp[l][k], s2 = dp[k][r];
        int p1 = s1.first + s2.first;
        long long p2 = s1.second + s2.second + s;
        if (p2 >= S) p2 = 0, p1++;
        dp[l][r] = max(dp[l][r], make_pair(p1, p2));
      }
    }
  }
  for (int i = 0; i < n; i++) {
    d = max(d, dp[i][nxt[i]].first);
  }
  return d;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++)
    scanf("%lld%lld", x + i, y + i), nxt[i] = (i + 1) % n;
  long long l = 0, r = (1ll << 60);
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (check(mid) >= k + 1)
      l = mid;
    else
      r = mid;
  }
  printf("%lld\n", l);
}

```