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;
ostream& operator<<(ostream& out, const vector<long long>& b) {
  for (auto k : b) out << k << " ";
  return out;
}
istream& operator>>(istream& in, pair<long long, long long>& b) {
  in >> b.first >> b.second;
  return in;
}
ostream& operator<<(ostream& out, const pair<long long, long long>& b) {
  out << "{" << b.first << ", " << b.second << "}";
  return out;
}
template <typename T1, typename T2>
inline bool chkmin(T1& x, const T2& y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool chkmax(T1& x, const T2& y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
const long long INF = 1e18;
const long double PI = acos(-1);
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long MAXN = 2001;
const long long K = 13;
long long n, x;
long long a[MAXN], t[K][MAXN];
long long pw[K];
long long ask(long long l, long long r) {
  long long lg = pw[r - l + 1];
  return min(t[lg][l], t[lg][r - (1 << lg) + 1]);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  pw[1] = 0;
  for (long long i = 2; i < K; i++) {
    pw[i] = pw[i / 2] + 1;
  }
  cin >> n >> x;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    t[0][i] = a[i];
  }
  for (long long i = 1; i < K; i++) {
    for (long long j = 0; j <= n - (1 << i); j++) {
      t[i][j] = min(t[i - 1][j], t[i - 1][j + (1 << (i - 1))]);
    }
  }
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    ans += a[i];
  }
  for (long long l = 1; l < n; l++) {
    long long cur = 0;
    for (long long j = 0; j < n; j++) {
      if (j < l) {
        cur += min(ask(n - (l - j), n - 1), ask(0, j));
      } else {
        cur += ask(j - l, j);
      }
    }
    chkmin(ans, cur + l * x);
  }
  cout << ans << "\n";
}
