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;
using ll = long long;
template <typename T>
string join(const vector<T>& v, string d = " ") {
  stringstream s;
  for (ll(i) = 0; (i) < ((ll)(v).size()); ++(i)) s << d << v[i];
  return s.str().substr(d.length());
}
template <typename T>
ostream& operator<<(ostream& s, const vector<T>& v) {
  if ((ll)(v).size()) s << join(v, ", ");
  return s;
}
const ll INF = 1e15;
const ll MOD = 1e9 + 7;
const double EPS = 1e-10;
int dy[] = {0, 0, 1, -1};
int dx[] = {1, -1, 0, 0};
struct edge {
  ll from, to, cost;
  edge(ll f, ll t, ll c) : from(f), to(t), cost(c) {}
  bool operator>(const edge& e) const { return cost > e.cost; }
};
int V;
vector<edge> G[2004];
ll d[2004];
bool used[2004];
ll X[2004];
void dijkstra(ll s) {
  priority_queue<edge, vector<edge>, greater<edge>> que;
  fill(d, d + V, INF);
  fill(used, used + V, false);
  fill(X, X + V, 0);
  d[s] = 0;
  for (ll(i) = 0; (i) < (G[s].size()); ++(i)) que.push(G[s][i]);
  while (!que.empty()) {
    edge e1 = que.top();
    que.pop();
    if (!used[e1.to]) {
      used[e1.to] = true;
      d[e1.to] = e1.cost;
      if (e1.from == 0) X[e1.to] = e1.cost;
      if (X[e1.from] != 0) X[e1.to] = X[e1.from];
      for (ll(i) = 0; (i) < (G[e1.to].size()); ++(i)) {
        edge e2 = G[e1.to][i];
        que.push(edge(e2.from, e2.to, e1.cost + e2.cost));
      }
    }
  }
}
int main() {
  ll N, x;
  cin >> N >> x;
  V = N + 1;
  for (ll(i) = 0; (i) < (N); ++(i)) {
    ll a;
    cin >> a;
    G[0].push_back(edge(0, i + 1, a));
    G[i + 1].push_back(edge(i + 1, (i + 1) % N + 1, x));
  }
  dijkstra(0);
  ll M = 0;
  ll MI = -1;
  for (ll(i) = 0; (i) < (V); ++(i))
    if (M < d[i]) M = d[i], MI = i;
  ll ans = M;
  for (ll(i) = 0; (i) < (V); ++(i))
    if (i != MI) ans += X[i];
  cout << ans << endl;
  return 0;
}
