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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

///#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
///#pragma GCC target("sse,sse2,sse3,ssse3,sse4,abm,mmx,popcnt,avx,avx2,tune=native")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>

/*
    zet:
        find_by_order(k): k-stat [k > 0]
        order_of_key(k) : {el < k}

    gp_hash_table<key, val, custom_hash>

    rope<T>:
        iterators: mutable_begin, ..
        methods  : erase, insert, substr, ..
*/

#define all(x) (x).begin(), (x).end()
#define size(x) (int)((x).size())
#define em_back emplace_back

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;

using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

template<typename T>
using zet = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

mt19937 eng(chrono::steady_clock::now().time_since_epoch().count());

const int N = 2e3 + 4;
const int INF = INT_MAX;

int n, x, a[N];
int used[N];

ll getMin(int state) {
    ll res = 0;
    int mn = INF;
    for (int i = 0; i < n; ++i) if (!used[i]) {
        mn = min(mn, a[i]);
    }
    for (int i = 0; i < n; ++i) if (!used[i]) {
        if (a[i] == mn) {
            used[i] = state;
            return mn;
        }
    }
}

void cycle(int dx) {
    if (dx == 1) {
        int saveUsed = used[n - 1];
        for (int i = n - 1; i > 0; --i) {
            used[i] = used[i - 1];
        }
        used[0] = saveUsed;
    } else {
        int saveUsed = used[0];
        for (int i = 0; i < n - 1; ++i) {
            used[i] = used[i + 1];
        }
        used[n - 1] = saveUsed;
    }
}

ll calc() {
    return getMin(0);
}

void solve() {
    cin >> n >> x;
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }

    ll res = 0;
    while (accumulate(used, used + n, 0) != n) {
        ll here = calc();

        cycle(1);
        ll to = x + calc();
        cycle(-1);

        if (to < here) {
            res += x;
            cycle(1);
        } else {
            res += getMin(1);
        }
    }

    cout << res << "\n";
}

int main() {
    ios::sync_with_stdio(0);
    cout.tie(0), cin.tie(0);

    int z = 1;
    /// cin >> z;

    while (z--) {
        solve();
    }

    return 0;
}