//k項までの{最小, 最大}が分かっていれば, k+1項までの{最小, 最大}を計算できる. {}の中が同じなので, 最小と最大を持ってDPできそう。
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

int INF = 1e+16;
int n;
int a[16];
int dp[17][2];

signed main() {
	int i, j;
	
	cin >> n;
	for (i = 0; i < n; i++) cin >> a[i];
	for (i = 0; i <= n; i++) {
		dp[i][0] = INF;
		dp[i][1] = -INF;
	}
	dp[0][0] = 0;
	
	for (i = 0; i < n; i++) {
		for (j = 0; j < 2; j++) {
			if (dp[i][j] == INF || dp[i][j] == -INF) continue;
			dp[i + 1][0] = min(dp[i + 1][0], dp[i][j] + a[i]);
			dp[i + 1][1] = max(dp[i + 1][1], dp[i][j] + a[i]);
			dp[i + 1][0] = min(dp[i + 1][0], dp[i][j] - a[i]);
			dp[i + 1][1] = max(dp[i + 1][1], dp[i][j] - a[i]);
			dp[i + 1][0] = min(dp[i + 1][0], dp[i][j] * a[i]);
			dp[i + 1][1] = max(dp[i + 1][1], dp[i][j] * a[i]);
			if (a[i] != 0) {
				dp[i + 1][0] = min(dp[i + 1][0], dp[i][j] / a[i]);
				dp[i + 1][1] = max(dp[i + 1][1], dp[i][j] / a[i]);
			}
		}
	}
	
	cout << dp[n][1] << endl;
	return 0;
}