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>
#define ALL(obj) begin(obj), end(obj)
using namespace std;
template <class T>
vector<T> make_vec(size_t a) {
    return vector<T>(a);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template <class T>
bool chmax(T &a, const T &b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}

using ll = long long;
using ull = unsigned long long;
using P = pair<int, int>;
const int INF = 2100100100;
const int MOD = 1e9 + 7;

ll N, M, K, Q, W, H, L, R, C;
string S;

int main() {
    ll x;
    cin >> N >> x;
    vector<ll> a(N);
    ll minid = 0;
    ll minval = 1e18;
    for (int i = 0; i < N; i++) {
        cin >> a.at(i);
        if (chmin(minval, a[i])) {
            minid = i;
        }
    }

    for (int i = 0; i < minid; i++) {
        a.push_back(a[i]);
    }

    vector<P> blck;  // id, cnt
    int id = minid;
    int cnt = 1;
    for (int i = minid + 1; i < (int)a.size(); i++) {
        if (a[id % N] + x * (i - id) < a[i]) {  // ずらす時
            cnt++;
        } else {
            blck.push_back(make_pair(id % N, cnt));
            cnt = 1;
            id = i;
        }
    }
    if (cnt != 0) blck.push_back(make_pair(id % N, cnt));

    ll ans = 0;
    int size = 0;
    for (auto i : blck) {
        //cout << i.first << " " << i.second << endl;
        ans += a[i.first] * i.second;
        chmax(size, i.second);
    }
    ans += (size - 1) * x;
    cout << ans << endl;

    return 0;
}
