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;

#define REP(var, n)  for (decltype(n) var = 0; var < (n); var++)
#define RREP(var, n)  for (auto var = n - 1; var != static_cast<decltype(var)>(-1); var--)
#define FOR(var, a, b)  for (auto var = (a); var < (b); var++)
#define RFOR(var, a, b)  for (auto var = b - 1; var != a; var--)

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;

#define all(c) begin(c),end(c)

template<typename T>
std::ostream &operator<<(std::ostream &out, const std::vector<T> &v) {
    if (!v.empty()) {
        out << '[';
        std::copy(v.begin(), v.end(), std::ostream_iterator<T>(out, ", "));
        out << "\b\b]";
    }
    return out;
}

const int IINF = 1e9 + 10;


auto solve() {
    int N, x;
    cin >> N >> x;
    vi a(N);

    ll num = 0, sum = 0;
    int smallest = IINF;
    for (auto &&e:a) {
        cin >> e;
        smallest = min(smallest, e);
        if (e <= x) {
            num++;
            sum += e;
        }
    }

    return sum + (ll)(N - num) * smallest + (ll)x * (N - num);
}

int main() {
    cout << solve() << endl;
    return 0;
}
