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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx")
using namespace std;
const long long maxn = 2e5 + 20;
ostream& operator<<(ostream& out, vector<long long>& v) {
  for (auto& x : v) out << x << ' ';
  return out;
}
ostream& operator<<(ostream& out, pair<long long, long long>& v) {
  out << v.first << ", " << v.second;
  return out;
}
istream& operator>>(istream& in, pair<long long, long long>& a) {
  in >> a.first >> a.second;
  return in;
}
const long long inf = (long long)2e9;
const long double pi = asin(1) * 2;
const long double eps = 1e-8;
const long long mod = (long long)1e9 + 7;
const long long ns = 97;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long n, x;
  cin >> n >> x;
  vector<long long> goo(n);
  for (auto& x : goo) cin >> x;
  vector<long long> a = goo;
  long long mx = 0;
  for (long long i = 0; i < n; ++i) {
    long long t = 0;
    for (long long j = 0; j < n; ++j) {
      if (i == j) continue;
      if (goo[i] > (i - j + n) % n * x + goo[j] ||
          goo[i] == (i - j + n) % n * x + goo[j] && (i - j + n) % n < t) {
        a[i] = goo[j];
        t = (i - j + n) % n;
      }
    }
    mx = max(mx, t);
    ;
    ;
  };
  long long ans = 0;
  for (long long i = 0; i < n; ++i) ans += a[i];
  ans += mx * x;
  cout << ans << '\n';
}
