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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (ll i = 0; i < n; i++)
#define FOR(i, a, b) for (ll i = a; i < b; i++)
#define len(v) ll(v.size())
#define fi first
#define se second

template <class T>
void cout_vec(const vector<T> &vec){
  for(auto itr:vec) cout<<itr<<' ';
  cout<<endl;
}

typedef pair<ll,ll> P;
const ll mod=1e9+7;
const ll inf=1e15;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll n,l;
  cin>>n>>l;
  deque<P> a(n);
  vector<ll> ans;
  ll sum=0;
  rep(i,n){
    cin>>a[i].fi;
    a[i].se=i+1;
    sum+=a[i].fi;
  }
  while(true){
    if(len(a)==1) break;
    if(sum<l){
      cout<<"Impossible"<<endl;
      return 0;
    }
    P now1=a.front(),now2=a.back();
    if(now1.fi<=now2.fi){
      sum-=now1.fi;
      ans.push_back(now1.se);
      a.pop_front();
    }else{
      sum-=now2.fi;
      ans.push_back(now2.se-1);
      a.pop_back();
    }
  }
  cout<<"Possible"<<endl;
  cout_vec(ans);
}