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 <queue>
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <string>
#include <vector>
#include <stdlib.h>
#include <map>
#include <stdio.h>
using namespace std;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for(int i = (a); i < (int)(b); i++)
typedef long long ll;
typedef unsigned long long ull;
typedef vector<ll> vint;
typedef vector<string> vstr;
typedef pair<ll, ll> pint;
unsigned long long inf = (1LL << 62);
ll mod = 998244353;
pint dx[4] = { {1, 0}, {-1, 0}, {0, 1}, {0, -1}};


ll N, x;
vector<ll> a;
ll vis[2000];
int main() {
	
	cin >> N >> x;
	a.resize(N);
	ll mi = inf;
	rep(i, N) {
		cin >> a[i];

		mi = min(a[i], mi);

	}
	

	ll sum = 0;
	rep(i, N) {
		if (a[i] == mi) {
			sum += mi;
			vis[i] = 1;
		}
	}

	bool flag = true;
	ll index = 0;
	vector<ll> q;
	while (flag == true) {
		flag = false;

		for (index = 0; index < N; index++) {
			if (index > 0 && vis[index] == 0 && vis[index - 1] == 1) {
				q.push_back(index);
				flag = true;
			}

			else if (index == 0 && vis[index] == 0 && vis[N - 1] == 1) {
				q.push_back(index);
				flag = true;
			}
		}

		ll s = 0;

		rep(i, q.size()) {
			s += a[q[i]];
			vis[q[i]] = 1;
		}

		sum += min(s, x + mi * q.size());

		while (!q.empty()) { q.pop_back(); }

	}


	cout << sum << endl;

	return 0;
}