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
#pragma once
#include <sstream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <iostream>
#include <utility>
#include <set>
#include <cctype>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <numeric>
#include <iomanip>
#define M_PI acos(-1)
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 1000000007;
const int MAX_V = 201;
int now = 0;

const int MAX = 200000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
struct edge {
	int t;
	ll cost;
	edge(int _t, ll _cost) : t(_t)
		, cost(_cost) {
	}
};
bool isPrime(int n) {

	if (n == 1) return false;
	for (ll i = 2; i * i <= n; i++)
	{
		if (n % i == 0) return false;
	}
	return true;
}
void Solve() {
	
	int n;
	ll x;
	cin >> n >> x;
	ll a[2010];
	for (size_t i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	ll res = 0;
	ll aval = 0;
	for (size_t i = 0; i < n; i++)
	{
		ll val = a[i];
		int iv = i + n;
		int in = i;
		for (size_t j = 0; j < n; j++)
		{
			if (i == j) continue;
			ll tv;
			if (j < i) {
				tv = min((i - j) * x + a[j], val);
			}
			else {
				tv = min((iv - j) * x + a[j], val);
			}
			if (tv < val) {
				in = j;
				val = tv;
			}
		}
		aval += a[in];
		res = max(val - a[in],res);
	}
		

	cout << (res + aval) << endl;
}
int main() {
	Solve();
	return 0;
}

