#include <bits/stdc++.h>
using namespace std;
int n, k, ans = 0;
int a[20], sum[20];
int dp[20][20]; //dp[i][j] 表示前 i 个数里插上 j 个乘号的最优解 
int main() {
	cin >> n >> k;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
		sum[i] = sum[i - 1] + a[i];
		dp[i][0] = sum[i];
		dp[1][i] = a[i];
	}
	for(int i = 2; i <= n; i++)
		for(int j = 1; j <= min(k, i - 1); j++)
			for(int k = j; k < i; k++)
				dp[i][j] = max(dp[i][j], dp[k][j - 1] * (sum[i] - sum[k]));
	cout << dp[n][k];
	return 0;
}