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;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
unsigned long long sttoi(std::string str) {
  unsigned long long ret;
  std::stringstream ss;
  ss << str;
  ss >> ret;
  return ret;
}
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  if (b > a) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int comb(int n, int k) {
  int ans = 1;
  for (int i = 0; i < min(k, n - k); i++) {
    ans *= n - i;
    ans /= i + 1;
  }
  return ans;
}
long long mod_pow(long long n, long long p, long long mod) {
  if (p == 0) return 1 % mod;
  if (p < 2) {
    return n % mod;
  } else {
    long long d = 2, value = (n % mod) * (n % mod) % mod;
    while (d * 2 < p) {
      d *= 2;
      value *= value;
      value %= mod;
    }
    value %= mod;
    return ((value % mod) * (mod_pow(n, p - d, mod) % mod)) % mod;
  }
}
void solve() {
  int n, x;
  cin >> n >> x;
  vector<int> v;
  int minimum[2000];
  fill(minimum, minimum + n, 100000000);
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    v.push_back(a);
  }
  int sum = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (j < i) {
        minimum[i] = min(v[j] + x * (i - j), minimum[i]);
      } else if (i < j) {
        minimum[i] = min(v[j] + x * (n - j + i - 1), minimum[i]);
      } else {
        minimum[i] = min(minimum[i], v[j]);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    sum += minimum[i];
  }
  cout << (sum) << endl;
  ;
}
int main() {
  solve();
  return 0;
}
