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>
#include<algorithm>
#include <assert.h>
using namespace std;
#define rep(a,b,c) for(int a=b;a<=c;a++)
#define ll long long 
#define sum(a,b) a=a+b
#define repn(a,b,c) for(int a = b ; a>=c ; a--)
#define mem(a,b) memset(a,b,sizeof(a))
#define sl(n) scanf("%lld", &n)
#define sf(n) scanf("%lf", &n)
#define si(n) scanf("%d", &n)
#define ss(n) scanf("%s", n)
#define pii pair <int, int>
#define pll pair <long long, long long >
#define plp pair <int, pll >
#define pb push_back

ll max(ll a, ll b){
    if(a > b) return a;
    return b;
}

ll min(ll a ,ll b){
    if(a>b)return b;
    return a;
}

//------------------------------------------------------------------------------------------------------------------------


int main(){
    
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    ll a[(int)1e5 * 2 + 5];

    ll n ,k;
    cin>>n>>k;

    ll sum = 0;

    vector<int>ans;

    rep(i,1,n){
        cin>>a[i];
        sum += a[i];
    }

    int l = 1;
    int r = n;

    int cnt = 0;

    while(l < r){

        if(sum < k){
            break;
        }

       // cout<<l<<" "<<r<<"::"<<sum<<" STEP::"<<cnt<<endl;

        if(a[l] < a[r]){
            cnt++;
            ans.push_back(l);
            
            sum -= a[l];
            l++;
           
        }else if(a[r] <= a[l]){
            cnt++;
            
            ans.push_back(r-1);
            sum -= a[r];

            r--;
           
        }


        //cout<<l<<" "<<r<<"::"<<sum<<" STEP::"<<cnt<<endl;

    }


    if(cnt < (n-1) ){
        cout<<"Impossible"<<endl;
    }else{
        cout<<"Possible"<<endl;
        int si =  ans.size()-1;

        rep(i,0,si){
            cout<<ans[i]<<endl;
        }
    }

    return 0;
}