The Fair Nut lives in n story house. a_i people live on the i-th floor of the house. Every person uses elevator twice a day: to get from the floor where he/she lives to the ground (first) floor and to get from the first floor to the floor where he/she lives, when he/she comes back home in the evening. 

It was decided that elevator, when it is not used, will stay on the x-th floor, but x hasn't been chosen yet. When a person needs to get from floor a to floor b, elevator follows the simple algorithm: 

  * Moves from the x-th floor (initially it stays on the x-th floor) to the a-th and takes the passenger. 
  * Moves from the a-th floor to the b-th floor and lets out the passenger (if a equals b, elevator just opens and closes the doors, but still comes to the floor from the x-th floor). 
  * Moves from the b-th floor back to the x-th. 

The elevator never transposes more than one person and always goes back to the floor x before transposing a next passenger. The elevator spends one unit of electricity to move between neighboring floors. So moving from the a-th floor to the b-th floor requires |a - b| units of electricity.

Your task is to help Nut to find the minimum number of electricity units, that it would be enough for one day, by choosing an optimal the x-th floor. Don't forget than elevator initially stays on the x-th floor. 

Input

The first line contains one integer n (1 ≤ n ≤ 100) — the number of floors.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 100) — the number of people on each floor.

Output

In a single line, print the answer to the problem — the minimum number of electricity units.

Examples

Input

3
0 2 1


Output

16

Input

2
1 1


Output

4

Note

In the first example, the answer can be achieved by choosing the second floor as the x-th floor. Each person from the second floor (there are two of them) would spend 4 units of electricity per day (2 to get down and 2 to get up), and one person from the third would spend 8 units of electricity per day (4 to get down and 4 to get up). 4 ⋅ 2 + 8 ⋅ 1 = 16.

In the second example, the answer can be achieved by choosing the first floor as the x-th floor.
/*------------------------------------------
    Author : Suyash Chavan
    Walchand College of Engineering, Sangli
--------------------------------------------*/
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#pragma GCC target ("sse4")

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

using namespace std;
using namespace __gnu_pbds;

#define ll                          long long
#define ld                          long double
#define endl                        "\n"
#define pb                          push_back
#define mp                          make_pair
#define eb                          emplace_back
#define fill(a,v)                   memset(a, v, sizeof a)
#define all(a)                      a.begin(), a.end()
#define rall(vec)                   vec.rbegin(),vec.rend()
#define sortv(a)                    sort(a.begin(),a.end())
#define sorta(a,n)                  sort(a,a+n)
#define set_bits(a)                 __builtin_popcount(a)
#define ordered_set                 tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define fast_map(mp)                mp.reserve(1024);\
                                    mp.max_load_factor(0.25)

template <typename T1, typename T2>
inline std::ostream& operator << (std::ostream& os, const std::pair<T1, T2>& p)
{
    return os << "(" << p.first << ", " << p.second << ")";
}
template<typename T>
inline std::ostream &operator << (std::ostream & os,const std::vector<T>& v)
{
    bool first = true;
    os << "[";
    for(unsigned int i = 0; i < v.size(); i++)
    {
        if(!first)
            os << ", ";
        os << v[i];
        first = false;
    }
    return os << "]";
}
template<typename T>
inline std::ostream &operator << (std::ostream & os,const std::set<T>& v)
{
    bool first = true;
    os << "[";
    for (typename std::set<T>::const_iterator ii = v.begin(); ii != v.end(); ++ii)
    {
        if(!first)
            os << ", ";
        os << *ii;
        first = false;
    }
    return os << "]";
}
template<typename T1, typename T2>
inline std::ostream &operator << (std::ostream & os,const std::map<T1, T2>& v)
{
    bool first = true;
    os << "[";
    for (typename std::map<T1, T2>::const_iterator ii = v.begin(); ii != v.end(); ++ii)
    {
        if(!first)
            os << ", ";
        os << *ii ;
        first = false;
    }
    return os << "]";
}

/*--------------------------- CODE STARTS FROM HERE---------------------------*/

int solve()
{
    ll n,s;
    cin>>n>>s;
    
    vector<ll> nums(n,0);
    
    for(ll i=0;i<n;i++)
        cin>>nums[i];
        
    ll tot = accumulate(nums.begin(),nums.end(),0);
    
    if(tot<s)
    {
        cout<<-1<<endl;
        return 0;
    }
    
    ll least = *min_element(nums.begin(),nums.end());
    ll got = 0;
    
    for(ll i=0;i<n;i++)
        got+=(nums[i]-least);
        
    if(got>=s)
    {
        cout<<least<<endl;
        return 0;
    }
    
    cout<<least-((s-got)/n)-1<<endl;
    
    return 0;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll t=1;
    //cin>>t;
    while(t--)
        solve();

    return 0;
}