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
#include <algorithm>
#include <iterator>
#include <vector>
#include <iomanip>
#include <iostream>
#include <stack>
#include <queue>
#include <list>
#include <utility>

#include <string>
#include <cstring>

#include <stdio.h>
#include <cassert>

// M_PI for pi
#define _USE_MATH_DEFINES
#include <math.h>

#define MAX 100000005

using namespace std;

int N;
longlong x;

long long A[2005];
long long B[2005];

void slove() {
}


int main(int argc, char* argv[]) {
    cin >> N >> x;

    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }


    long long S = 1000000000000000000LL;
    long long s;


    for (int i = 0; i < N; ++i) {
        B[i] = A[i];
    }

    for (int k = 0; k < N; ++k) {
        for (int i = 0; i < N; ++i) {
            B[i] = min(B[i], A[(N + i - k) % N]);
        }

        //for (int i = 0; i < N; ++i) {
        //    cout << B[i] << " ";
        //}
        //cout << endl;

        s = 0;

        for (int i = 0; i < N; ++i) {
            s += B[i];
        }

        s += k * x;
        //cout << s << endl;

        S = min(s, S);
    }

    cout << S << endl;


    return 0;
}
