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 <cmath>
#include <string>
#include <algorithm>
#include <iomanip>
#include <map>
#include <bitset>
#include <cstdio>
#include <set>
#include <stack>
#include <queue>
#include <cassert>
#include <numeric>
#define rep(i,n) Rep(i,0,n)
#define Rep(i,k,n) for(int i=k ; i<n ; i++)
#define rep1(i,n) for(int i=1 ; i<=n ; i++)
#define vi vector<int>
#define vii vector<vector<int>>
#define mii map<int,int>
#define Sort(v) sort(v.begin(),v.end())
#define Reverse(v) reverse(v.begin(),v.end())
#define ALL(a)  (a).begin(),(a).end()
#define pb push_back
#define mp make_pair
//#define int ll
typedef long long ll;

const int md = 1000000007;
const int INF = 1<<30;
using namespace std;

void canP(bool x){
    if(x){
        cout << "Possible" << endl;
    }else{
        cout << "Impossible" << endl;
    }
}
int main(){
    ll n,l; cin >> n >> l;
    vi a(n);
    vector<pair<int,int> > k(n-1);
    ll sum = 0;
    
    rep(i,n){
        cin >> a[i];
        sum += a[i];
    }

    rep(i,n-1){
        k[i].first = a[i] + a[i+1];
        k[i].second = i+1;
    }

    Sort(k);
    bool ans = true;
    if(k[n-2].first < l) ans = false;

    canP(ans);
    if(ans){
        rep(i,n-1){
            cout << k[i].second << endl;
        }
    }
}