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 <cstdio>
#include <queue>
#include <vector>

using namespace std;

const int N = 100100;

int val[N];


int main(){
	int n;
	long long l;
	scanf("%d%lld",&n,&l);
	long long size = 0;
	for(int i = 0; i < n; i++){
		scanf("%d",&val[i]);
	}
	vector<int> ans;
	for(int i = 0; i < n-1; i++){
		if(val[i]+val[i+1]>=l){
			for(int j = 0; j < i; j++){
				ans.push_back(j+1);
			}
			for(int j = n-2; j >= i+1; j--){
				ans.push_back(j+1);
			}	
			ans.push_back(i+1);	
			break;
		}
	}	
	if(ans.size()>0){
		puts("Possible");
		for(auto a : ans)printf("%d ",a);
	}
	else{
		puts("Impossible");
	}


}