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)

#include "bits/stdc++.h"

#define REP(i,n) for(ll i=0;i<ll(n);++i)
#define RREP(i,n) for(ll i=ll(n)-1;i>=0;--i)
#define FOR(i,m,n) for(ll i=m;i<ll(n);++i)
#define RFOR(i,m,n) for(ll i=ll(n)-1;i>=ll(m);--i)
#define ALL(v) (v).begin(),(v).end()
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());
#define DUMP(v) REP(aa, (v).size()) { cout << v[aa]; if (aa != v.size() - 1)cout << " "; else cout << endl; }
#define INF 1000000001ll
#define MOD 1000000007ll
#define EPS 1e-9

const int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[8] = { 0,1,1,1,0,-1,-1,-1 };


using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll max(ll a, int b) { return max(a, ll(b)); }
ll max(int a, ll b) { return max(ll(a), b); }
ll min(ll a, int b) { return min(a, ll(b)); }
ll min(int a, ll b) { return min(ll(a), b); }


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	ll n, x;
	cin >> n >> x;
	vl v(n);
	REP(i, n)cin >> v[i];
	vvi q(n);
	REP(i, n) {
		int p = -1;
		ll cmin = INF*INF;
		REP(j, n) {
			if (cmin > v[(i + j) % n] + j * x) {
				cmin = v[(i + j) % n] + j * x;
				p = (i + j)%n;
			}
		}
		q[(i - p + n) % n].push_back(p);
	}
	
	ll ans = 0, cnt = 0;
	RREP(i, n) {
		if (cnt != 0)ans += x;
		REP(j, q[i].size()) {
			cnt++;
			ans += v[q[i][j]];
		}
	}
	cout << ans << endl;
}