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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cstdio>
#include <queue>
#include "math.h"
#include <complex>
#include <iomanip>
#define ifor(i,a,b) for (int i=(a);i<(b);i++)
#define rfor(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define rep(i,n) for (int i=0;i<(n);i++)
#define rrep(i,n) for (int i=(n)-1;i>=0;i--)
using namespace std;
typedef long double ld;
typedef long long int  lli;
typedef complex <double> P;
const double eps = 1e-11;
int vex[4]={1,0,-1,0};
int vey[4]={0,1,0,-1};
int main(){
	int N,L;
	cin >> N >> L;
	int x[100005];
	int y[100005];
	rep(i,N)cin >> x[i];
	bool flag = true;
	for(int i = 0;i+1<N;i++){
		y[i]=x[i]+x[i+1];
	}
	for(int i = 0;i+1<N;i++){
		if(y[i]<L&&y[i+1]<L)flag = false;
	}
	if(!flag){
		cout << "Impossible"<<endl;
		return  0;
	}
	cout << "Possible"<<endl;
	int cnt = 0;
	for(int i = 0;i+2<N;i=i+2){
		if(y[i]<L&&y[i+1]>=L){
			cout <<i+1<<endl;
			cout << i+2<<endl;
		}
		else {
			cout << i +2 << endl;
			cout << i +1 << endl;
		}
	
		
	}
	return 0;
	
}