Problem:
Problem Statement

We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i.

At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly:

Choose a (connected) rope with a total length of at least L, then untie one of its knots.

Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.

Constraints

2≤N≤10^5

1≤L≤10^9

1≤a_i≤10^9

All input values are integers.

Input

The input is given from Standard Input in the following format:

N L
a_1 a_2 ... a_n

Output

If it is not possible to untie all of the N-1 knots, print Impossible.

If it is possible to untie all of the knots, print Possible, then print another N-1 lines that describe a possible order to untie the knots. The j-th of those N-1 lines should contain the index of the knot that is untied in the j-th operation. Here, the index of the knot connecting piece i and piece i+1 is i.

If there is more than one solution, output any.

Sample Input 1

3 50
30 20 10

Sample Output 1

Possible
2
1

If the knot 1 is untied first, the knot 2 will become impossible to untie.

Sample Input 2

2 21
10 10

Sample Output 2

Impossible

Sample Input 3

5 50
10 20 30 40 50

Sample Output 3

Possible
1
2
3
4

Another example of a possible solution is 3, 4, 1, 2.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

            #include<bits/stdc++.h>
            #define ll long long
            #define f   first
            #define s   second
            #define pb          push_back
            #define mod        1000000007
            #define mod1       1000000009
            #define hell        998244353
            #define inf         1000000000000000000LL
            #define pi          3.14159265358979323
            #define N           1000050
            using namespace std;
            ll n,k,sum,q,res,m,val;
            ll x,y;
            void solve()    
            {   

                cin>>n>>m;
                vector<ll> v(n);
                for(int i=0;i<n;i++) cin>>v[i];
                for(int i=0;i<n-1;i++)
                {
                    if(v[i]+v[i+1]>=m)
                    {   cout<<"Possible\n";
                        for(int j=0;j<i;j++)
                            cout<<j+1<<"\n";
                        
                        for(int j=n-1;j>=(i+1);j--)
                            cout<<j<<"\n";
                        return;
                    }
                }
                cout<<"Impossible";









            }
            int main()
            {   ios_base::sync_with_stdio(false);
                cin.tie(0);cout.tie(0);
                ll t=1;
                //cin>>t;
                for(int i=1;i<=t;i++)
                    solve();
                

            }