Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 10
1 100

Sample Output 1

12

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.

Catch a slime in color 1. This takes 1 second.

Sample Input 2

3 10
100 1 100

Sample Output 2

23

Snuke can act as follows:

Catch a slime in color 2. This takes 1 second.

Cast the spell. The color of the slime changes: 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Cast the soell. The color of each slime changes: 3 → 1, 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Sample Input 3

4 10
1 2 3 4

Sample Output 3

10

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Catch a slime in color 2. This takes 2 seconds.

Catch a slime in color 3. This takes 3 seconds.

Catch a slime in color 4. This takes 4 seconds.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) for(int i=0;i<(n);i++)
#define RREP(i,n) for(int i=(n)-1;i>=0;i--)
using namespace std;
int N, x;

int inc(int i){
    if(i == N - 1) return 0;
    else return i + 1;
}

int dec(int i){
    if(i == 0) return N - 1;
    else return i - 1;
}

int main()
{
    scanf("%d %d", &N, &x);
    vector<int> a(N);
    int m = 99999999, index = 0;;
    REP(i, N){
        scanf("%d", &a[i]);
        if(m > a[i]){
            m = a[i];
            index = i;
        }
    }
    vector<int> p(N), call(N), num(N);
    REP(i, N){
        call[i] = 0;
        num[i] = 0;
    }
    p[index] = a[index];
    call[index] = 1;
    int i = inc(index);
    while(1){
        if(i == index) break;
        if(p[dec(i)] + x <= a[i]){
            num[i] = num[dec(i)] + 1;
            p[i] = p[dec(i)] + x;
            int j = i;
            while(1){
                j = dec(j);
                if(call[j] != 0){
                    call[j]++;
                    break;
                }
            }
        }else{
            call[i]++;
            p[i] = a[i];
        }
        i = inc(i);
    }
    long long ans = 0;
    int M = 0;
    REP(i, N){
        ans += call[i] * a[i];
        M = max(num[i], M);
        //printf("call[%d] = %d\n", i, call[i]);
        //printf("num[%d] = %d\n", i, num[i]);
    }
    ans += M * x;
    printf("%lld\n", ans);
}