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: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(),v.end()
#define _GLIBCXX_DEBUG
#define fi first
#define se second

using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;
using vd = vector<double>;
using vvi = vector<vi>;
using vvll = vector<vll>;
using vvd = vector<vd>;
using vvvi = vector<vvi>;
using vvvll = vector<vvll>;
using vvvd = vector<vvd>;

const double pi=acos(-1);
const ll MOD=1e9+7;
const ll INF=(1LL<<60);

int main() {
    ll n,l;
    cin>>n>>l;
    ll a[n];
    rep(i,n)cin>>a[i];
    int index=-1;
    rep(i,n-1){
        if (a[i]+a[i+1]>=l){
            index=i;
            break;
        }
    }
    if (index==-1){
        cout<<"Impossible"<<endl;
        return 0;
    }
    cout<<"Possible"<<endl;
    for (int i=1;i<index+1;i++)cout<<i<<endl;
    for (int i=n-1;i>index+1;i--)cout<<i<<endl;
    cout<<index+1<<endl;

    return 0;
}