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.

Examples

Input

2 10
1 100


Output

12


Input

3 10
100 1 100


Output

23


Input

4 10
1 2 3 4


Output

10
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<string>
#include<map>
#include<cmath>
#include<queue>	
#include<functional>
#include<climits>
#include<tuple>
#include<utility>
#include<cassert>

using namespace std;

//define
#define ALL(a) a.begin(),a.end()
#define REP(i,n) for(int i=0;i<n;i++)
#define RREP(i,n) for(int i=n-1;i>=0;i--)
#define debug(x) if(1)cout<<#x<<":"<<x<<endl;
#define DEBUG(x) if(1)cout<<#x<<":"<<x<<endl;
#define ll long long
#define pb push_back
#define PB push_back
#define lb lower_bound
#define LB lower_bound
#define ub upper_bound
#define UB upper_bound


//constant
const int MOD = 1000000007;
const int INF = INT_MAX / 3 - 1;
const double EPS = 1e-9;
const int dx4[4] = { 1,0,-1,0 };
const int dy4[4] = { 0,1,0,-1 };
const int dx8[8] = { -1,0,1,-1,1,-1,0,1 };
const int dy8[8] = { -1,-1,-1,0,0,1,1,1 };

//typedef
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef pair<int, int> pii;


int dat[2*2000];
int x = 3e9;

int n;

void segment_tree(int n_, int x) {
	n = 1;
	while (n < n_) n *= 2;
	REP(i, 2 * n - 1) dat[i] = x;
}


int F(int p, int q) {//function

	return min(p, q);
}

void update(int k, int a) {
	k += n - 1;
	dat[k] = a;

	while (k > 0) {
		k = (k - 1) / 2;
		dat[k] = F(dat[k * 2 + 2], dat[k * 2 + 1]);
	}
}


int query(int a, int b, int k, int l, int r) {
	if (r <= a || b <= l) return INF;
	if (a <= l&&r <= b) return dat[k];
	else {
		int v1 = query(a, b, k * 2 + 1, l, (l + r) / 2);
		int v2 = query(a, b, k * 2 + 2, (l + r) / 2, r);
		return F(v1, v2);
	}

}

int main() {
	int n;
	int x;
	cin >> n >> x;
	vi a(n);
	REP(i, n)
		cin >> a[i];
	segment_tree(n, INF);
	REP(i, n)
		update(i, a[i]);
	long long int res = 1e15;
	REP(i, n) {
		long long int sum = 0;
		REP(j, n)
			if (j - i < 0)
				sum += min(query(0, j + 1, 0, 0, n), query(n + j - i, n, 0, 0, n));
			else
				sum += query(j - i, j + 1, 0, 0, n);
		res = min(res, sum + long long int(x)*i);

	}
	cout << res << endl;


	return 0;
}