Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 10
1 100

Sample Output 1

12

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Cast the spell. The color of the slime changes: 1 → 2. This takes 10 seconds.

Catch a slime in color 1. This takes 1 second.

Sample Input 2

3 10
100 1 100

Sample Output 2

23

Snuke can act as follows:

Catch a slime in color 2. This takes 1 second.

Cast the spell. The color of the slime changes: 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Cast the soell. The color of each slime changes: 3 → 1, 2 → 3. This takes 10 seconds.

Catch a slime in color 2. This takes 1 second.

Sample Input 3

4 10
1 2 3 4

Sample Output 3

10

Snuke can act as follows:

Catch a slime in color 1. This takes 1 second.

Catch a slime in color 2. This takes 2 seconds.

Catch a slime in color 3. This takes 3 seconds.

Catch a slime in color 4. This takes 4 seconds.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

long long int dp[2010][2010];		// 0~i回変化させてj番目にするために掛かる時間の最小値
long long int N, change_count[2010];			// j番目を作るときに魔法で変化した回数
long long int min(long long int a, long long int b) {return a < b ? a : b;}
//void print_dp(void);

int main(void)
{
	int i, j;
	long long int x, a[2010], max = 0, total = 0;
	long long int ans = 0;

	scanf("%lld%lld", &N, &x);
	for (i = 0; i < N; i++) {
		scanf("%lld", &a[i]);
		// 直接捕まえる場合
		dp[0][i] = a[i];
		change_count[i] = 0;
	}

	for (i = 1; i < N; i++) {
		for (j = 0; j < N; j++) {
			dp[i][j] = min(dp[i - 1][j], dp[i - 1][(N + j - i) % N] + x * i);
			if (dp[i][j] != dp[i - 1][j])
				change_count[j] = change_count[(N + j - i) % N] + i; 
		}
	}

	for (i = 0; i < N; i++) {
		ans += dp[N - 1][i];
		max = max > change_count[i] ? max : change_count[i];
		total += change_count[i];
	}
	ans -= (total - max) * x;
	//print_dp();

	printf("%lld\n", ans);
	return 0;
}

/*void print_dp(void) 
{
	int i, j;
	for (i = 0; i < N; i++) {
		for (j = 0; j < N; j++) {
			printf("%lld(%d) ", dp[i][j], change_count[j]);
		}
		putchar('\n');
	}
}*/