Problem:
Score : 100 points

Problem Statement

There are N stones, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), the height of Stone i is h_i.
Here, h_1 < h_2 < \cdots < h_N holds.

There is a frog who is initially on Stone 1.
He will repeat the following action some number of times to reach Stone N:

If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, N. Here, a cost of (h_j - h_i)^2 + C is incurred, where j is the stone to land on.

Find the minimum possible total cost incurred before the frog reaches Stone N.

Constraints

All values in input are integers.

2 \leq N \leq 2 \times 10^5

1 \leq C \leq 10^{12}

1 \leq h_1 < h_2 < \cdots < h_N \leq 10^6

Input

Input is given from Standard Input in the following format:

N C
h_1 h_2 \ldots h_N

Output

Print the minimum possible total cost incurred.

Sample Input 1

5 6
1 2 3 4 5

Sample Output 1

20

If we follow the path 1 → 3 → 5, the total cost incurred would be ((3 - 1)^2 + 6) + ((5 - 3)^2 + 6) = 20.

Sample Input 2

2 1000000000000
500000 1000000

Sample Output 2

1250000000000

The answer may not fit into a 32-bit integer type.

Sample Input 3

8 5
1 3 4 5 10 11 12 13

Sample Output 3

62

If we follow the path 1 → 2 → 4 → 5 → 8, the total cost incurred would be ((3 - 1)^2 + 5) + ((5 - 3)^2 + 5) + ((10 - 5)^2 + 5) + ((13 - 10)^2 + 5) = 62.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
 
#define X first
#define Y second
 
using namespace std;
 
typedef long long ll;
typedef pair<ll, ll> pii;
 
const int N = 2e5 + 50;
 
int n;
ll c;
ll h[N];
ll dp[N];
vector<int> env;
 
ll _ccw(pii a, pii b, pii c) {
  return a.X * (b.Y - c.Y) + b.X * (c.Y - a.Y) + c.X * (a.Y - b.Y);
}
 
int ccw(pii a, pii b, pii c) {
  ll get = _ccw(a, b, c);
 
  if(get > 0)
    return 1;
  else if (get < 0)
    return -1;
  return get;
}
 
pii conv(int i) {
  return {-2LL * h[i], h[i] * h[i] + dp[i]};
}
 
void insert(int i) {
  if(env.empty()) {
    env.push_back(i);
    return;
  }
 
  int top = env.back();
  env.pop_back();
 
  while(!env.empty() && ccw(conv(env.back()), conv(top), conv(i)) >= 0)
    top = env.back(), env.pop_back();
 
  env.push_back(top);
  env.push_back(i);
}
 
ll comp(int i, ll val) {
  return -2LL * h[i] * val + h[i] * h[i] + dp[i];
}
 
int query(ll val) {
  int lo = 0, hi = env.size() - 1, mid;
 
  while(lo < hi) {
    mid = (lo + hi) >> 1;
  
    if(comp(mid, val) >= comp(mid + 1, val))
      lo = mid + 1;
    else
      hi = mid;
  }
 
  return lo;
}
 
void load() {
  scanf("%d%lld",  &n, &c);
 
  for(int i = 0; i < n; ++i)
    scanf("%lld", h + i);
}
 
void solve () {
  dp[0] = 0;
 
  insert(0);
 
  for(int i = 1; i < n; ++i) {
    int j = query(h[i]);
    dp[i] = (h[i] - h[j]) * (h[i] - h[j]) + c + dp[j];
    assert(dp[i] >= 0);
    insert(i);
  }
 
  printf("%lld\n", dp[n - 1]);
}
 
int main() {
  load();
  solve();
 
  return 0;
}