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 <cstdio>
#include <iostream>
#include <bitset>
#include <cmath>
#include <vector>
#include <algorithm>
#include <string>
 
using namespace std;
typedef long long ll;
 
#define ALL(obj) (obj).begin(), (obj).end() 
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb(e) push_back(e)
#define mp(a, b) make_pair(a, b)

 int main(){
    int n,l,f=-1,s;
    cin >> n >> l;
    vector<int> a(n); 
    rep(i,n) cin >> a[i];
    rep(i,n-1){
        if(a[i]+a[i+1] >= l){
            f = i; s = i+1;
            break;
        }
    }
    if(f==-1){cout << "Impossible" << endl;}
    else{
        cout << "Possible" << endl;
        rep(i,f) cout << i+1 << endl;
        FOR(i,s,n-1) cout << i+1 << endl;
        cout << f+1 << endl;
    }

    return 0;
 }