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 int ll ;

long long gcd(long long aaa,long long bbb) {
    if(bbb==0) {
        return aaa;
    }
    return gcd(bbb,aaa%bbb);
}

int main (){

    ll N,x;
    cin >> N >> x;
    ll a[N],s=0,cmp,tmp[2000]={},l=0,count=-1,o,countt=0,n[2000]={},tmpp[2000]={};
    map<ll,ll>m;
    map<ll,ll>c;

    for(int i = 0; i < N; i++) {
        tmpp[i] = 1e9;
    }

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

    for(int i = 0; i < N; i++) {
        tmp[i] = a[i];
        for(int j = 0; j < N; j++) {
            cmp = i + 1;
            cmp += j;
            if(cmp >= N) {
                cmp -= N;
            }
            if(i == cmp) {
                break;
            }
            tmp[i] += min(a[cmp], a[i] + x);
            if(min(a[cmp],a[i] + x) == a[cmp]) {
                m[cmp] = 1;
            }
        }
    }

    for(int i = 0; i < N*2; i++) {
        if(i == N && count == 0) {
            break;
        }else if(i == N) {
            countt++;
        }
        o = i % N;
        if(count == -1) {
            if(m[i] == 1) {
                l = i;
                count++;
            }
        }else {
            if(m[i] == 1) {
                n[count] = abs(o - l) * (a[l] + x);
                count++;
                l = o;
                if(countt == 1) {
                    break;
                }
            }
        }
    }

    sort(n,n+2000);

    ll ans = s;

    countt = count+1;
    if(count >= 2) {
        for(int i = 0; i < N; i++) {
            if(m[i] == 1) {
                for(int j = 1; j <= count; j++) {
                    tmpp[j] += n[count-j];
                }
                tmpp[i] *=
            }else {
                break;
            }
        }
    }
    for(int i = 0; i < N; i++) {
        tmp[i] = min(tmp[i],tmpp[i]);
    }
    for(int i = 0; i < N; i++) {
        ans = min(tmp[i],ans);
    }

    cout << ans << endl;
    return 0;
}
