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;
long long N, X;
long long A[2010];
struct edge {
  long long to, cost;
};
typedef pair<long long, long long> P;
long long V;
vector<edge> G[2010];
long long d[2010];
long long prev_p[2010];
void dijkstra(long long s) {
  priority_queue<P, vector<P>, greater<P> > que;
  fill(d, d + V, 1000000000);
  fill(prev_p, prev_p + V, -1);
  d[s] = 0;
  que.push(P(0, s));
  while (!que.empty()) {
    P p = que.top();
    que.pop();
    long long v = p.second;
    if (d[v] < p.first) {
      continue;
    }
    for (long long i = 0; i < G[v].size(); ++i) {
      edge e = G[v][i];
      if (d[e.to] > d[v] + e.cost) {
        d[e.to] = d[v] + e.cost;
        prev_p[e.to] = v;
        que.push(P(d[e.to], e.to));
      }
    }
  }
}
vector<long long> get_path(long long t) {
  vector<long long> path;
  for (; t != -1; t = prev_p[t]) {
    path.push_back(t);
  }
  reverse(path.begin(), path.end());
  return path;
}
int main() {
  cin >> N >> X;
  V = N + 1;
  for (int i = 1; i <= N; ++i) {
    long long a;
    cin >> a;
    A[i] = a;
    edge e;
    e.to = i;
    e.cost = a;
    G[0].push_back(e);
    for (int j = 1; j <= N; ++j) {
      if (i == j) {
        continue;
      }
      e.to = j;
      e.cost = X;
      G[i].push_back(e);
    }
  }
  dijkstra(0);
  long long ans = 0;
  for (int i = 1; i <= N; ++i) {
    ans += d[i];
  }
  printf("%lld\n", ans);
  return 0;
}
