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 = 1 << 30;
const double EPS = 1e-10;
double zero(double d) { return d < EPS ? 0.0 : d; }
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << '(' << p.first << ',' << p.second << ')';
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &a) {
  os << '[';
  for (int i = 0; i < (int)(a.size()); i++) os << (i ? " " : "") << a[i];
  return os << ']';
}
string toString(int i) {
  stringstream ss;
  ss << i;
  return ss.str();
}
const int MOD = 1000000007;
long long fpow(long long a, long long k, int M) {
  long long res = 1ll;
  long long x = a;
  while (k != 0) {
    if ((k & 1) == 1) res = (res * x) % M;
    x = (x * x) % M;
    k >>= 1;
  }
  return res;
}
struct prepare {
  prepare() {
    cout.setf(ios::fixed, ios::floatfield);
    cout.precision(8);
    ios_base::sync_with_stdio(false);
  }
} _prepare;
int main() {
  long long N, x;
  cin >> N >> x;
  vector<long long> a(N);
  for (int i = 0; i < (int)(N); i++) cin >> a[i];
  const long long INF = 1ll << 60;
  vector<pair<long long, int> > cost(N, make_pair(INF, 0));
  for (int i = 0; i < (int)(N); i++) {
    for (int j = 0; j < (int)(N); j++) {
      chmin(cost[i], make_pair(a[(i - j + N) % N] + (long long)x * j, -j));
    }
  }
  long long ans = 0;
  long long ma = 0;
  for (int i = 0; i < (int)(N); i++) {
    ans += cost[i].first - (long long)cost[i].second * -1 * x;
    chmax(ma, (long long)cost[i].second * -1);
  }
  cout << ans + ma * x << endl;
  return 0;
}
