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 <vector>
#include <algorithm>
#include <cmath>
#include <set>
#include <queue>
#include <deque>
#include <map>
#include <stack>
#include<bitset>
#include<list>
#include<cassert>
#include<numeric>
 
using namespace std;

const int N = 2 * 1000 + 5;
vector<long long> a;
long long n, x;
long long dp[N][N];

int main()
{
	cin >> n >> x;
	for (int i = 0; i < n; i++)
	{
		int tmp;
		cin >> tmp;
		a.push_back(tmp);
	}
	long long ans = 1000 * 1000 * 1000;
	for (int i = 0; i < n; i++)
		dp[0][i] = a[i];
	for (int i = 1; i < n; i++)
	{
		for (int j = 0; j < n; j++)
		{
			dp[i][j] = min(a[((i - k) + n) % n], dp[(i - 1 + n) % n][j]); 
		}
	}
	for (int i = 0; i < n; i++)
	{
		long long tmp = 0;
		for (int j = 0; j < n; j++)
		{
			tmp += dp[i][j];
		}
		ans = min(ans, tmp);
	}
	cout << ans << endl;

}