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;
template <class C>
void min_self(C &a, C b) {
  a = min(a, b);
}
template <class C>
void max_self(C &a, C b) {
  a = max(a, b);
}
const long long MOD = 1000000007;
long long mod(long long n, long long m = MOD) {
  n %= m, n += m, n %= m;
  return n;
}
const int MAXN = 1e5 + 5;
const int LOGN = 21;
const long long INF = 1e14;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T1, class T2, class T3>
void add(T1 &x, T2 y, T3 m = MOD) {
  x += y;
  if (x >= m) x -= m;
}
template <class T1, class T2, class T3>
void sub(T1 &x, T2 y, T3 m = MOD) {
  x -= y;
  if (x < 0) x += m;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n), pref(n + 1);
  long long mn = INF;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mn = min(mn, a[i]);
    pref[i + 1] = pref[i] + a[i];
  }
  auto sum = [&](int l, int r) -> long long {
    if (l > r) return 0;
    return pref[r + 1] - pref[l];
  };
  vector<bool> ok(n);
  long long ans = mn * n;
  while (1) {
    bool f = 0;
    for (int i = 0; i < n; i++) {
      if (ok[i] == 0 && a[i] == mn) {
        ok[i] = 1;
        f = 1;
      }
    }
    if (f == 0) break;
    int cnt = 0;
    for (int i = 0; i < n; i++) {
      if (ok[i]) cnt++;
    }
    if (cnt == n) break;
    ans += x;
    vector<bool> nok(n);
    for (int i = 1; i < n; i++) nok[i] = ok[i - 1];
    nok[0] = ok[n - 1];
    ok = nok;
  }
  ans = min(ans, pref[n]);
  cout << ans, cout << "\n";
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
