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<iostream>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<string.h>
#include<queue>
#include<stack>
#include<climits>
#include<cmath>
using namespace std;

int main(){
    int n,l;
    int a[100001];
    vector<int>b;
    cin>>n>>l;
    for(int i=0;i<n;i++){
        cin>>a[i];
        if ((i>0&&a[i]+a[i-1]<l&&n==2)||(i>1&&a[i]+a[i-1]<l&&a[i-1]+a[i-2]<l)){
            cout<<"Impossible"<<endl;
            return 0;
        } else if(i>0&&a[i]+a[i-1]<l){
            b.push_back(i);
        }
    }
    cout<<"Possible"<<endl;
    for(int i=0;i<b.size();i++){
        cout<<b[i]<<endl;
    }
    int c=0;
    for(int i=1;i<n;i++){
        if(i==b[c]){
            c++;
        } else {
            cout<<i<<endl;
        }
    }
    
    return 0;
}