#include <iostream>
#include <cmath>
using namespace std;
 
#define MAX 100001
 
int main() {
	int amount_of_pltf;
	int heights[MAX], energy[MAX], platforms[MAX], list_of_pltf[MAX];
	cin >> amount_of_pltf;
	for (int i = 1; i <= amount_of_pltf; i++) cin >> heights[i];
 
	energy[1] = 0;
	energy[2] = abs(heights[2] - heights[1]);
	platforms[1] = -1;
	platforms[2] = 1;
 
	//В цикле вычисляем значения ячеек energy[j] и platforms[j](3 ≤ j ≤ amount_of_pltf)
	for (int j = 3; j <= amount_of_pltf; j++){
		if((energy[j-1] + abs(heights[j] - heights[j-1])) < (energy[j-2] + 3 * abs(heights[j] - heights[j-2]))){
			energy[j] = energy[j-1] + abs(heights[j] - heights[j-1]);
			platforms[j] = j - 1;
		} else {
			energy[j] = energy[j-2] + 3 * abs(heights[j] - heights[j-2]);
			platforms[j] = j - 2;
		}
	}
 
	//В переменной final_amount вычисляем конечное количество платформ, по которым нужно пройти.
	int final_amount = 0;
	for (int l = amount_of_pltf; l > 0; l = platforms[l]) list_of_pltf[final_amount++] = l;
 
	//Выводим результат. Минимальное количество энергии,
	//которое следует потратить для попадания с первой платформы на последнюю, равно energy[amount_of_pltf].
	cout << energy[amount_of_pltf] << "\n" << final_amount << "\n";
	for(int k = final_amount - 1; k >= 0; k--) cout << list_of_pltf[k] << " ";
	return 0;
}

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