Problem:
Score : 700 points

Problem Statement

You have an integer sequence of length N: a_1, a_2, ..., a_N.

You repeatedly perform the following operation until the length of the sequence becomes 1:

First, choose an element of the sequence.

If that element is at either end of the sequence, delete the element.

If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.

You would like to maximize the final element that remains in the sequence.

Find the maximum possible value of the final element, and the way to achieve it.

Constraints

All input values are integers.

2 \leq N \leq 1000

|a_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

In the first line, print the maximum possible value of the final element in the sequence.

In the second line, print the number of operations that you perform.

In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.

If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.

Sample Input 1

5
1 4 3 7 5

Sample Output 1

11
3
1
4
2

The sequence would change as follows:

After the first operation: 4, 3, 7, 5

After the second operation: 4, 3, 7

After the third operation: 11(4+7)

Sample Input 2

4
100 100 -1 100

Sample Output 2

200
2
3
1

After the first operation: 100, 200(100+100)

After the second operation: 200

Sample Input 3

6
-1 -2 -3 1 2 3

Sample Output 3

4
3
2
1
2

After the first operation: -4, 1, 2, 3

After the second operation: 1, 2, 3

After the third operation: 4

Sample Input 4

9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 4

5000000000
4
2
2
2
2

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma GCC optimize ("O3")
#pragma GCC target ("avx")
#include "bits/stdc++.h" // define macro "/D__MAI"

using namespace std;
typedef long long int ll;

#define debug(v) {printf("L%d %s > ",__LINE__,#v);cout<<(v)<<endl;}
#define debugv(v) {printf("L%d %s > ",__LINE__,#v);for(auto e:(v)){cout<<e<<" ";}cout<<endl;}
#define debuga(m,w) {printf("L%d %s > ",__LINE__,#m);for(int x=0;x<(w);x++){cout<<(m)[x]<<" ";}cout<<endl;}
#define debugaa(m,h,w) {printf("L%d %s >\n",__LINE__,#m);for(int y=0;y<(h);y++){for(int x=0;x<(w);x++){cout<<(m)[y][x]<<" ";}cout<<endl;}}
#define ALL(v) (v).begin(),(v).end()
#define repeat(cnt,l) for(auto cnt=0ll;(cnt)<(l);++(cnt))
#define rrepeat(cnt,l) for(auto cnt=(l)-1;0<=(cnt);--(cnt))
#define iterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);++(cnt))
#define diterate(cnt,b,e) for(auto cnt=(b);(cnt)!=(e);--(cnt))
#define MD 1000000007ll
#define PI 3.1415926535897932384626433832795
template<typename T1, typename T2> ostream& operator <<(ostream &o, const pair<T1, T2> p) { o << "(" << p.first << ":" << p.second << ")"; return o; }
template<typename T> T& maxset(T& to, const T& val) { return to = max(to, val); }
template<typename T> T& minset(T& to, const T& val) { return to = min(to, val); }
void bye(string s, int code = 0) { cout << s << endl; exit(code); }
mt19937_64 randdev(8901016);
inline ll rand_range(ll l, ll h) {
    return uniform_int_distribution<ll>(l, h)(randdev);
}

#if defined(_WIN32) || defined(_WIN64)
#define getchar_unlocked _getchar_nolock
#define putchar_unlocked _putchar_nolock
#elif defined(__GNUC__)
#else
#define getchar_unlocked getchar
#define putchar_unlocked putchar
#endif
namespace {
#define isvisiblechar(c) (0x21<=(c)&&(c)<=0x7E)
    class MaiScanner {
    public:
        template<typename T> void input_integer(T& var) {
            var = 0; T sign = 1;
            int cc = getchar_unlocked();
            for (; cc<'0' || '9'<cc; cc = getchar_unlocked())
                if (cc == '-') sign = -1;
            for (; '0' <= cc && cc <= '9'; cc = getchar_unlocked())
                var = (var << 3) + (var << 1) + cc - '0';
            var = var * sign;
        }
        inline int c() { return getchar_unlocked(); }
        inline MaiScanner& operator>>(int& var) { input_integer<int>(var); return *this; }
        inline MaiScanner& operator>>(long long& var) { input_integer<long long>(var); return *this; }
        inline MaiScanner& operator>>(string& var) {
            int cc = getchar_unlocked();
            for (; !isvisiblechar(cc); cc = getchar_unlocked());
            for (; isvisiblechar(cc); cc = getchar_unlocked())
                var.push_back(cc);
            return *this;
        }
        template<typename IT> void in(IT begin, IT end) { for (auto it = begin; it != end; ++it) *this >> *it; }
    };
    class MaiPrinter {
    public:
        template<typename T>
        void output_integer(T var) {
            if (var == 0) { putchar_unlocked('0'); return; }
            if (var < 0)
                putchar_unlocked('-'),
                var = -var;
            char stack[32]; int stack_p = 0;
            while (var)
                stack[stack_p++] = '0' + (var % 10),
                var /= 10;
            while (stack_p)
                putchar_unlocked(stack[--stack_p]);
        }
        inline MaiPrinter& operator<<(char c) { putchar_unlocked(c); return *this; }
        inline MaiPrinter& operator<<(int var) { output_integer<int>(var); return *this; }
        inline MaiPrinter& operator<<(long long var) { output_integer<long long>(var); return *this; }
        inline MaiPrinter& operator<<(char* str_p) { while (*str_p) putchar_unlocked(*(str_p++)); return *this; }
        inline MaiPrinter& operator<<(const string& str) {
            const char* p = str.c_str();
            const char* l = p + str.size();
            while (p < l) putchar_unlocked(*p++);
            return *this;
        }
        template<typename IT> void join(IT begin, IT end, char sep = '\n') { for (auto it = begin; it != end; ++it) *this << *it << sep; }
    };
}
MaiScanner scanner;
MaiPrinter printer;






ll m, n, kei;
ll aa[1010];

void special() {
    auto p = max_element(aa, aa + n);
    int mini = p - aa;

    printer << *p << '\n' << n - 1 << '\n';

    repeat(i, mini)
        printer << (1) << '\n';

    repeat(i, n-mini-1)
        printer << (2) << '\n';

}

int main() {
    scanner >> n;
    scanner.in(aa, aa + n);

    // todo: 全てがマイナス

    ll best = 0;

    vector<int> answer_pre;
    vector<int> answer;

    int choiceOdd = 0;
    {
        ll odd, even;
        odd = even = 0;
        repeat(i, n) {
            (i % 2 ? odd : even) += max(0ll, aa[i]);
        }
        choiceOdd = odd >= even;
        best = max(odd, even);
    }

    if (best == 0) {
        special();
        return 0;
    }


    int left = 0, right = n - 1, offset = 0;

    while (true) {
        if (left % 2 == choiceOdd && aa[left] > 0) {
            break;
        }
        answer_pre.push_back(left + offset + 1);
        offset -= 1;
        left += 1;
    }

    while (true) {
        if (right % 2 == choiceOdd && aa[right] > 0) {
            break;
        }
        right -= 1;
    }
    
    for (int p = right + 1; p < n; ++p){
        answer_pre.push_back(right + offset + 1 + 1);
    }
    
    {
        bool first = true;
        iterate(i, left, right+1) {
            if (i % 2 != choiceOdd) continue;
            if (aa[i] <= 0) {
                answer_pre.push_back(i + offset + 1);
                offset -= 2;
            }
            else {
                if (!first)
                    answer.push_back(i + offset - 1 + 1),
                    offset -= 2;
                first = false;
            }
        }
    }


    printer << best << '\n';

    printer << ll(answer.size() + answer_pre.size()) << '\n';

    for (auto e : answer_pre)
        printer << e << '\n';

    for (auto e : answer)
        printer << e << '\n';


    return 0;
}