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
// -----------------------------------
// author   : MatsuTaku
// country  : Japan
// created  : 09/03/20 01:08:28
// -----------------------------------

#include <bits/stdc++.h>
#include "../../../library/include/segment_tree.hpp"

using namespace std;
using ll = long long;

struct Min {
  int x;
  Min(int x=1e9): x(x) {}
  Min operator*(const Min& r) const {return min(x,r.x);}
  Min& operator*=(const Min& r) {return *this = *this * r;}
};
using Rmq = SegmentTree<Min>;

int main() {
  cin.tie(nullptr); ios::sync_with_stdio(false);

  int n,x; cin>>n>>x;
  vector<int> A(n); for (auto& a:A) cin>>a;
  Rmq rmq(A.begin(), A.end());

  vector<long long> dp(n);
  {
    for (int i = 0; i < n; i++) {
      dp[i] = A[(n-i)%n];
      if (i > 0)
        dp[i] = min(dp[i], dp[i-1]);
    }
  }
  constexpr ll INF = 1e18;
  for (int i = 1; i < n; i++) {
    auto nxt = dp;
    for (int j = 0; j < n; j++) {
      auto mina = rmq.query(max(i-j, 0), i+1).x;
      if (i-j < 0)
        mina = min(mina, rmq.query(i-j+n, n).x);
      nxt[j] = dp[j] + mina;
      if (j > 0)
        nxt[j] = min(nxt[j], nxt[j-1]);
    }
    dp = nxt;
  }
  ll ans = INF;
  for (int i = 0; i < n; i++) {
    ans = min(ans, dp[i] + (ll)x * i);
  }
  cout << ans << endl;

  return 0;
}
