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;
long long const MOD = 1e9 + 7;
int const inf = 1e9;
long long const INF = 1e18;
inline void ios_() {
  cin.tie(0);
  ios::sync_with_stdio(false);
}
template <typename T>
int size(const T& a) {
  return (int)a.size();
}
template <typename T>
T Div(T a, T b) {
  return (a + b - 1) / b;
}
template <typename T>
bool chmin(T& a, const T& b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
signed main() {
  int n;
  long long x;
  cin >> n >> x;
  vector<long long> v(n);
  for (auto& i : v) cin >> i;
  int p = min_element((v).begin(), (v).end()) - v.begin();
  long long magic = 0, cnt = 0;
  long long m = v[p], ans = 0;
  for (int(i) = 0; (i) < (n); (i)++) {
    if (v[(p + 1) % n] <= m + x) {
      ans += m * (cnt + 1);
      m = v[(p + 1) % n];
      chmax(magic, cnt);
      cnt = 0;
    } else {
      cnt++;
    }
    cerr << "ans"
         << " = " << ans << endl;
    p = (p + 1) % n;
  }
  ans += magic * x;
  cout << ans << endl;
  return 0;
}
