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: Wrong Answer)

/*Arnab's Code
*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define vll vector<ll>
#define vii vector<int>
#define pll pair<ll,ll>
#define pii pair<int,int>
#define take(a,n) for(ll i=0;i<n;i++)cin>>a[i]
#define takes(a,n,st) for(ll i=st;i<n+st;i++)cin>>a[i]
ll sum(ll a[],ll n,ll s){ for(ll i=0;i<n;i++)s+=a[i]; return s;}
#define mod 1000000009
ll power(ll x, ll y, ll p)  {  
    ll res = 1;  
    x = x % p;   
    if (x == 0) return 0;  
    while (y > 0){  
        if (y & 1)  
            res = (res*x) % p;  
        y = y>>1;
        x = (x*x) % p;  
    }  
    return res;  
}
ll a[100017];
void solve(){
	ll n,l,mini,i,x,minp;
	cin>>n>>l;
	take(a,n);
	mini=-1;
	for(i=1;i<n;i++){
		x=a[i]+a[i-1];
		if(x>=l and x>mini){
			mini=x;
			minp=i;
		}
	}
	if(mini == -1){
		cout<<"Impossible";
		return;
	}
	cout<<"Possible\n";
	for(i=1;i<n;i++){
		if(i!=minp)
			cout<<i<<"\n";
	}
	cout<<minp;
	return;
}
int main(){
//	ios::sync_with_stdio(false);
//	cin.tie(NULL);
//	#ifndef ONLINE_JUDGE
//		freopen("input.txt","r",stdin);
//		freopen("output.txt","w",stdout);
//	#endif
	ll t,T;
//	cin>>T;
	T=1;
	for(t=1;t<=T;t++){
//		cout<<"Case #"<<t<<": ";
		solve();
	}
	return 0;
}