Problem:
Score : 400 points

Problem Statement

Snuke lives in another world, where slimes are real creatures and kept by some people.
Slimes come in N colors. Those colors are conveniently numbered 1 through N.
Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him a_i seconds.

Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N-1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.

Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

2≤N≤2,000

a_i are integers.

1≤a_i≤10^9

x is an integer.

1≤x≤10^9

Input

The input is given from Standard Input in the following format:

N x
a_1 a_2 ... a_N

Output

Find the minimum time that Snuke needs to have slimes in all N colors.

Sample Input 1

2 10
1 100

Sample Output 1

12

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.

Catch a slime in color 1. This takes 1 second.

Sample Input 2

3 10
100 1 100

Sample Output 2

23

Snuke can act as follows:

Catch a slime in color 2. This takes 1 second.

Cast the spell. The color of the slime changes: 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Cast the soell. The color of each slime changes: 3 → 1, 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Sample Input 3

4 10
1 2 3 4

Sample Output 3

10

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Catch a slime in color 2. This takes 2 seconds.

Catch a slime in color 3. This takes 3 seconds.

Catch a slime in color 4. This takes 4 seconds.

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

#define _CRT_SECURE_NO_WARNINGS

#include <bits/stdc++.h>
using namespace std;

#define F first
#define S second

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> llll;
typedef pair<int, ii> iii;
typedef vector<ii> vii;
typedef vector<ll> vll;
typedef vector<llll> vllll;
typedef vector<int> vi;
typedef vector<ull> vull;

// TIMER
std::clock_t __start;
double __duration;
void start_timer() { __start = std::clock(); }
void print_timer() {
	__duration = (std::clock() - __start) / (double)CLOCKS_PER_SEC;
	std::cout << "Duration (msec): " << __duration * 1000 << '\n';
}
// END TIMER

//PI
# define M_PI		3.14159265358979323846
// fast input
#define scan(x) do{while((x=getchar())<'0'); for(x-='0'; '0'<=(_=getchar()); x=(x<<3)+(x<<1)+_-'0');}while(0)
//char _;

template <class T>
T min_(T a, T b) { return (a < b ? a : b); }
template <class T>
T max_(T a, T b) { return (a > b ? a : b); }
double EPS = 1e-9;
bool eq_(const double& lhs, const double &rhs) {
	return (fabs(lhs - rhs) < EPS);
}

const int INF = int(2e9);
const ll INF_LL = LLONG_MAX;


vi prices;
int N, X;
vector<vi> DP;
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> N >> X;
	prices.resize(N);
	for (int i = 0; i < N; ++i) {
		cin >> prices[i];
	}
	DP.assign(N + 1, vi(N + 1, 0));
	for (int i = 0; i < N; ++i) {
		DP[i][i] = prices[i];
	}
	for (int len = 1; len < N; ++len) {
		for (int L = 0; L + len < N; ++L) {
			int R = L + len;
			vi tries = { 
				DP[L][R - 1] + prices[R], 
				prices[L] + DP[L + 1][R],
				DP[L][R - 1] + prices[L] + X, 
				DP[L + 1][R] + prices[R] + X,
				DP[L][R - 1] + prices[R - 1] + X, 
				DP[L + 1][R] + prices[L + 1] + X
			};
			sort(tries.begin(), tries.end());
			DP[L][R] = tries[0];
		}
	}
	cout << DP[0][N - 1] << '\n';
}