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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define pb push_back
#define ppb pop_back
#define fi first
#define se second
#define mid ((x + y) / 2)
#define left (ind * 2)
#define right (ind * 2 + 1)
#define mp make_pair
#define timer ((double)clock() / CLOCKS_PER_SEC)
#define endl "\n"
#define spc " "
#define d1(x) cerr<<#x<<":"<<x<<endl
#define d2(x, y) cerr<<#x<<":"<<x<<" "<<#y<<":"<<y<<endl
#define d3(x, y, z) cerr<<#x<<":"<<x<<" "<<#y<<":"<<y<<" "<<#z<<":"<<z<<endl
#define fast_io() ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;

typedef long long int lli;
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<double, double> dd;

const int N = (int)(2e3 + 5);
const int LOG = (int)(20);

lli n, cs, ar[N], mini[N][N], tree[N], ans = INT_MAX;

void build(int ind, int x, int y) {
	if(x == y) {
		tree[ind] = ar[x];
		return;
	}
	build(left, x, mid);
	build(right, mid + 1, y);
	tree[ind] = min(tree[left], tree[right]);
}

lli get(int ind, int x, int y, int a, int b) {
	if(a > b)
		return min(get(ind, x, y, 1, b), get(ind, x, y, a, n));
	if(y < a || b < x)
		return INT_MAX;
	if(a <= x && y <= b)
		return tree[ind];
	return min(get(left, x, mid, a, b), get(right, mid + 1, y, a, b));
}

int main() {
	fast_io();
	//freopen("inp.in", "r", stdin);

	cin >> n >> cs;
	for(int i = 1; i <= n; i++)
		cin >> ar[i];

	build(1, 1, n);

	for(int k = 0; k <= n; k++) {
		lli t = k * cs;
		for(int i = 1; i <= n; i++) {
			int l = ((i - k) < 1) ? i - k + n : i - k;
			int r = i;
			d2(l, r);
			t += get(1, 1, n, l, r);
		}
		ans = min(ans, t);
	}
	cout << ans;
}