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.

Examples

Input

5 6
1 2 3 4 5


Output

20


Input

2 1000000000000
500000 1000000


Output

1250000000000


Input

8 5
1 3 4 5 10 11 12 13


Output

62
#include <bits/stdc++.h>

using namespace std;

const int MAXN = 200005;

long long h[MAXN], dp[MAXN];

long double f(int i, int j)
{
    return (dp[i] - dp[j] + h[i]*h[i] - h[j]*h[j]) / (2.0 * (h[i]-h[j]));
}

int main()
{
    int n;
    long long c;
    scanf("%i%lli", &n, &c);
    for (int x = 1; x <= n; x++)
        scanf("%lli", &h[x]);
    deque<int> buf;
    buf.push_back(1);
    for (int x = 2; x <= n; x++)
    {
        while (buf.size() >= 2 && f(buf[0], buf[1]) <= h[x])
            buf.pop_front();
        dp[x] = dp[buf.front()] + (h[x] - h[buf.front()]) * (h[x] - h[buf.front()]) + c;
        while (buf.size() >= 2 && f(buf[buf.size()-2], buf[buf.size()-1]) >= f(buf[buf.size()-1], x))
            buf.pop_back();
        buf.push_back(x);
    }
    printf("%lli\n", dp[n]);
}
