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<vector>
#include<algorithm>
#include<string>
#include<queue>

using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<=n;++i)

signed main(){
    int n,l;cin>>n>>l;
    vector<int> a(n);
    rep(i,n)cin>>a[i];
    bool flag=true;
    int sum=0;
    rep(i,n){
        sum+=a[i];
    }
    vector<int> res;
    int left=1;int right=n-1;
    while(right-left>0&&sum>=l){
        if(a[left-1]==a[right]){
            int bufl=left;int bufr=right;
            while(bufl<=bufr&&a[bufl-1]==a[bufr]){
                bufl++;bufr--;
                //if(bufl>bufr)break;
            }
            if(a[bufl-1]<a[bufr]){
                for(int i=left;i<bufl;i++){
                    res.push_back(i);
                    sum-=a[i-1];
                    left=bufl;
                }
            }
            else{
                for(int i=right;i>bufr;i--){
                    res.push_back(i);
                    sum-=a[i];
                    right=bufr;
                }
            }
            if(sum<l)break;
        }
        else if(a[left-1]>a[right]){
            res.push_back(right);
            sum-=a[right];
            right--;
        }
        else{
            res.push_back(left);
            sum-=a[left-1];
            left++;
        }
    }
    if(sum>=l){
        res.push_back(right);
    }
    int cnt=res.size();
    if(cnt!=n-1){
        cout<<"Impossible"<<endl;
    }
    else{
        cout<<"Possible"<<endl;
        rep(i,cnt){
            cout<<res[i]<<endl;
        }
    }
    //cout<<cnt<<endl;
    return 0;
}