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<string>
#include<cmath>
#include<algorithm>
#include<cctype>
#include<queue>
#include<deque>
#include<regex>
#include<stack>
#include<stdio.h>
#include<vector>
#include<set>
#include<map>
#include<iomanip>

#define rep(i, n) for(int i=0;i<n;i++)

typedef int long long ll;
using namespace std;
typedef pair<int, int> P;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
const ll MOD = 1e9 + 7;
static const int MAX = 100;
static const int INF = (1 << 23);

int main() {
    ll n,l;
    cin>>n>>l;
    vector<ll> himo(n);
    rep(i,n){
        cin>>himo[i];

    }
    int ans=0;
    bool  f = false;
    for(int i=0;i<n-1;i++){
        
        ll k=himo[i]+himo[i+1];
        if(k>=l){
            ans=i+1;
            f=true;
        }
    }

    if(f) {
        cout<<"Possible"<<endl;
        for (int i = 1; i < n; i++) {
            if (i != ans) {
                cout << i << endl;
            }
        }
        cout << ans << endl;
    }else{

        cout<<"Impossible"<<endl;
    }

    return 0;
}