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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <stdio.h>
#include <bitset>
using namespace std;
#define FOR(I,A,B) for(int I = (A); I < (B); ++I)
typedef long long ll;
const ll inf = 1e15;
int main()
{
  int N;
  cin>>N;
  ll x;
  cin>>x;
  ll a[N];
  FOR(i,0,N) cin >> a[i];
  int from[N]; // どこから来るか
  FOR(i,0,N) from[i] = i;
  ll mx = 0;
  FOR(i,0,N)
  {
    ll mnmn = inf;
    int cnt = 0;
    int j = i;
    while(cnt < N)
    {
      if(mnmn > a[j] + cnt * x)
      {
        mnmn = a[j] + cnt * x;
        mx = max(mx, cnt);
        from[i] = j;
      }
      j += N-1;
      j %= N;
      cnt++;
    }
  }
  ll ans = mx * x;
  FOR(i,0,N)
  {
    ans += a[from[i]];
  }
  cout << ans << endl;
  return 0;
}