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;
const static int MOD = 1000000000 + 7;
const static long long int INF = 1LL << 60;
const static int dy[] = {0, 1, 0, -1};
const static int dx[] = {1, 0, -1, 0};
template <typename T>
struct SparseTable {
  vector<vector<T> > st;
  vector<int> lookup;
  SparseTable(const vector<T> &v) {
    int b = 0;
    while ((1 << b) <= v.size()) ++b;
    st.assign(b, vector<T>(1 << b));
    for (int i = 0; i < v.size(); i++) {
      st[0][i] = v[i];
    }
    for (int i = 1; i < b; i++) {
      for (int j = 0; j + (1 << i) <= (1 << b); j++) {
        st[i][j] = min(st[i - 1][j], st[i - 1][j + (1 << (i - 1))]);
      }
    }
    lookup.resize(v.size() + 1);
    for (int i = 2; i < lookup.size(); i++) {
      lookup[i] = lookup[i >> 1] + 1;
    }
  }
  inline T rmq(int l, int r) {
    int b = lookup[r - l];
    return min(st[b][l], st[b][r - (1 << b)]);
  }
};
signed main(int argc, char *argv[]) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;
  long long int x;
  cin >> N >> x;
  vector<long long int> slime(2 * N);
  for (int i = 0; i < N; ++i) {
    long long int a;
    cin >> a;
    slime[i] = a;
    slime[i + N] = a;
  }
  SparseTable<long long int> st(slime);
  priority_queue<long long int, vector<long long int>, greater<long long int> >
      PQ;
  for (int k = 0; k < N; ++k) {
    long long int ans = 0;
    for (int i = 0; i < N; ++i) {
      ans += st.rmq(N + i - k, N + i + 1) + k * x;
    }
    PQ.push(ans);
  }
  cout << PQ.top() << endl;
  return 0;
}
