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 <utility>
using namespace std;
typedef pair<long long, long long> P;
long long MAX = 1e18;
struct segK{
    int n;
    vector<P> dat;
    void init(int n_){
        n = 1;
        while(n < n_) n *= 2;
        dat.resize(n * 2 - 1);
        for(int i = 0; i < n * 2 - 1; i++) dat[i] = P(MAX, 0);
    }
    void update(int k, long long p, long long c){
        k += n - 1;
        dat[k] = P(p, c);
        while(k > 0){
            k = (k - 1) / 2;
            if(dat[2 * k + 1].first < dat[2 * k + 2].first) dat[k] = dat[2 * k + 1];
            else if(dat[2 * k + 1].first >  dat[2 * k + 2].first) dat[k] = dat[2 * k + 2];
            else if(dat[2 * k + 1].second < dat[2 * k + 2].second) dat[k] = dat[2 * k + 1];
            else dat[k] = dat[2 * k + 2];
        }
    }
    P query(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return P(MAX, 0);
        if(l <= a && b <= r) return dat[k];
        else{
            P v1 = query(a, b, k * 2 + 1, l, (l + r) / 2);
            P v2 = query(a, b, k * 2 + 2, (l + r) / 2, r);
            if(v1.first > v2.first) return v2;
            else if(v1.first < v2.first) return v1;
            else if(v1.second < v2.second) return v1;
            else return v2;
        }
    }
};
int main(){
    long long N;
    long long x;
    cin >> N >> x;
    vector<segK> segtree(N);
    vector<long long> a(N);
    for(int i = 0; i < N; i++) {
        cin >> a[i];
        segtree[i].init(N);
    }
    for(int i = 0; i < N; i++){
        for(long long j = 0; j < N; j++){
            segtree[i].update(j, a[(i + j) % N] + (N - j) % N * x, (N - j) % N);
        }
    }
    long long ans = 0;
    vector<P> vec;
    long long cnt = 0;
    for(int i = 0; i < N; i++){
        P temp = segtree[i].query(0, N, 0, 0, segT.n);
        vec.push_back(temp);
        cnt = max(cnt, temp.second);
    }
    for(int i = 0; i < N; i++) ans += vec[i].first - x * vec[i].second;
    ans += cnt * x;
    cout << ans << endl;
}
