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<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
#include<utility>
#include<numeric>
#include<vector>
#include<map>
#include<unordered_map>
#include<set>
#include<tuple>
#include<stack>
#include<queue>
#include<functional>
#include<iterator>
#include<cmath>
#include<cctype>
#include<cassert>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;

const int INF = 1e9;
const ll LINF = 1e18;
const int MOD = INF+7;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define rep(i,n) FOR(i,0,n)

struct edge{int to,cost;};

void dump(ll *b,int n){
	for(int i=0;i<n;i++){
		cout << b[i] << " ";
	}
	cout << "\n";
}

ll mypow(ll x,ll n){
	if(n==0) return 1;
	ll res=mypow(x*x,n/2);
	if(n&1) res=res*x;
	return res;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	
	int n;
	ll x;
	cin >> n >> x;
	ll a[n];
	rep(i,n) cin >> a[i];

	int time[n]={};
	ll cost[n]={};
	
	cost[0]=a[0];
	FOR(i,1,n){
		if(a[i]<=cost[i-1]+x){
			cost[i]=a[i];
		}
		else{
			cost[i]=cost[i-1]+x;
			time[i]=time[i]+1;
		}
	}
	bool updated=false;
	if(a[0] > cost[n-1]+x){
		cost[0]=cost[n-1]+x;
		time[0]=time[n-1]+1;
		updated=true;
	}
	int cur=1,pre=0;
	int step=2000;
	while(true){
		if(cost[cur]>cost[pre]+x){
			cost[cur]=cost[pre]+x;
			time[cur]=time[pre]+1;
			cur++; pre++;
			if(cur==n) cur=0;
			else if(pre==n) pre=0;
		}else break;
	}
	int mt=0;
	int s;
	ll ans=0;
	rep(i,n){
		s=i-time[i];
		if(s<0) s+=n;
		ans+=a[s];
		mt = max(mt,time[i]);
	}
	cout << ans+mt*x << "\n";
	
	return 0;
}