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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
using namespace std;
#define int long long
const int N=5002020;
int n,L,sum,top,a[N],ans[N];
int read(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
#undef int
int main()
{
	#define int long long
	n=read(); L=read();
	for(int i=1;i<=n;i++)
		a[i]=read(),sum+=a[i];
	int l=1,r=n;
	while(l<r && sum>=L){
		if(a[l]<a[r])sum-=a[l],ans[++top]=l,++l;
		else sum-=a[r],--r,ans[++top]=r;
	}
	if(l+1<r || (r==l+1 && sum<L))puts("Impossible");
	else{
		puts("Possible");
		for(int i=1;i<n;i++)printf("%lld\n",ans[i]);
	}
	return 0;
}