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 = numeric_limits<int>::max() / 4;
const long long linf = numeric_limits<long long>::max() / 4;
const unsigned long long ulinf = numeric_limits<unsigned long long>::max() / 4;
const double pi = acos(-1);
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const int dx8[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
const int dy8[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
long long modpow(long long i, long long j) {
  long long tmp = 1;
  while (j) {
    if (j % 2) tmp = ((tmp * i) % (long long)(1e9 + 7));
    i = ((i * i) % (long long)(1e9 + 7));
    j /= 2;
  }
  return tmp;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
vector<string> split(const string &str, char sep) {
  vector<string> v;
  stringstream ss(str);
  string buffer;
  while (getline(ss, buffer, sep)) {
    v.push_back(buffer);
  }
  return v;
}
int main() {
  long long n, x;
  long long a[3000] = {};
  int mi = 0;
  cin >> n >> x;
  for (int i = (0); i < (int)(n); i++) {
    cin >> a[i];
    if (a[mi] > a[i]) mi = i;
  }
  long long ans2 = linf;
  for (int j = (0); j < (int)(n); j++) {
    mi = j;
    long long ans = 0;
    for (int i = (0); i < (int)(n); i++) {
      int num = (mi + i) % n;
      ans += min(a[num], x + a[mi]);
    }
    ans2 = min(ans2, ans);
  }
  cout << (ans2) << endl;
  return 0;
}
