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<bits/stdc++.h>
using namespace std;
typedef long long LL;

const int maxn = 2e3 + 5;
LL a[maxn];
LL b[maxn][maxn];//b[i][k] : 当整个过程使用了【k】次魔法，获得【color i】时的最少花费的时间。(不包括【k * x】).

int main(){
    int n;
    LL x,sum = 1LL << 60LL;
    cin >> n >> x;
    for(int i = 1;i <= n;i++) cin >> a[i];
    for(int i = 1;i <= n;i++){
        b[i][0] = a[i];
        for(int k = 1;k <= n - 1;k++){
            int pos = i - k;
            if(pos <= 0) pos = n - abs(pos);
            b[i][k] = min(b[i][k - 1],a[pos]);//【b[i][k]】的转移方程.
        }
    }
    for(int k = 0;k <= n - 1;k++){
        LL ans = 0;
        for(int i = 1;i <= n;i++){
            ans += b[i][k];
        }
        sum = min(sum,ans + x * k);//取最小值.
    }
    cout << sum << endl;
}1

