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 int inf = 0x7fffffff;
const long long INF = 0x7fffffffffffffff;
template <typename A, typename B>
ostream &operator<<(ostream &s, const pair<A, B> &p) {
  return s << "(" << p.first << "," << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &s, const vector<T> &c) {
  s << "[ ";
  for (auto it : c) s << it << " ";
  s << "]";
  return s;
}
template <typename T>
ostream &operator<<(ostream &o, const set<T> &st) {
  o << "{";
  for (auto it = st.begin(); it != st.end(); it++)
    o << (it == st.begin() ? "" : ", ") << *it;
  return o << "}";
}
template <typename T1, typename T2>
ostream &operator<<(ostream &o, const map<T1, T2> &make_pair) {
  o << "{";
  for (auto it = make_pair.begin(); it != make_pair.end(); it++) {
    o << (it == make_pair.begin() ? "" : ", ") << it->first << ":"
      << it->second;
  }
  o << "}";
  return o;
}
long long n, x, a[2005];
int main() {
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
  }
  long long maxx = 0;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long cur_ans = a[i], cur_x = 0;
    long long tmp = a[i];
    for (int j = (i + 1) % n; j != i; j = (j + 1) % n) {
      int step = (j - i + n) % n;
      if (cur_ans > a[j] + step * x ||
          cur_ans == a[j] + step * x && step > cur_x)
        cur_ans = a[j] + step * x, tmp = a[j], cur_x = step;
    }
    ans += tmp;
    maxx = max(maxx, cur_x);
  }
  cout << ans + x * maxx << endl;
  return 0;
}
