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;
struct UnionFind {
  vector<int> par;
  UnionFind(int n) : par(n, -1) {}
  void init(int n) { par.assign(n, -1); }
  int root(int x) {
    if (par[x] < 0)
      return x;
    else
      return par[x] = root(par[x]);
  }
  bool issame(int x, int y) { return root(x) == root(y); }
  bool merge(int x, int y) {
    x = root(x);
    y = root(y);
    if (x == y) return false;
    if (par[x] > par[y]) swap(x, y);
    par[x] += par[y];
    par[y] = x;
    return true;
  }
  int size(int x) { return -par[root(x)]; }
};
int gcd(int a, int b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  long long x;
  cin >> x;
  vector<long long> a(n), mn(n), s(n);
  int mx = 0;
  for (int(i) = 0; (i) < (n); ++(i)) {
    cin >> a[i];
    mn[i] = a[i];
    s[i] = a[i];
  }
  for (int(i) = 0; (i) < (n); ++(i)) {
    for (int j = 1; j < n; ++j) {
      if (chmin(mn[(i + j) % n], mn[i] + j * x)) {
        s[(i + j) % n] = s[i];
        chmax(mx, j);
      }
    }
  }
  long long res = x * mx;
  for (int(i) = 0; (i) < (n); ++(i)) res += s[i];
  cout << res << endl;
}
