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 <cstdio>
#include <iostream>
#include <cmath>
#include <cstring> 
#include <sstream>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <memory.h>
#include <iomanip>
#include <bitset>
#include <list>
#include <stack>
 
using namespace std;
 
#define pai 3.14159265359
#define mod 1000000007

long long int b[2001][2001];

int main()
{
    long long int n, x;
    cin >> n >> x;
    long long int a[2001];
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    for(int i = 1; i <= n; i++){
        b[i][0] = a[i];
    }
    for(int i = 1; i <= n; i++){
        for(int j = 1; j < n; j++){
            int moto = i - j;
            if(moto < 1) moto += n;
            b[i][j] = min(b[i][j - 1], a[moto]);
        }
    }
    // b[i][j]は魔法j回以下つかったときにスライムiを得るためのスライムを捕まえるのにかかる最短時間
    long long int ans = (long long int)pow(mod, 2);
    for(int i = 0; i < n; i++){
        // iは魔法を使う回数
        long long int result = x * i;
        for(int j = 1; j <= n; j++){
            result += b[j][i];
        }
        ans = min(ans, result);
    }
    cout << ans << endl;
    return 0;
}v