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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

#define MAX_N 100010

int main(void) {
	int n,L,a[MAX_N],num[MAX_N];
	int i,judge;
	
	scanf("%d %d", &n,&L);
	
	i=0;
	while(i<n) {
		scanf("%d", &a[i]);
		i++;
	}
	
	i=0;
	judge=1;
	while(i<=n-2) {
		if (a[i]+a[i+1]<L && a[i+1]+a[i+2]<L) {
			judge=-1;
			break;
		}
		
		if (a[i]+a[i+1]>=L && a[i+1]+a[i+2]>=L) {
			num[i]=i+1;	num[i+1]=i+2;
			i=i+2;
		}
		else if (a[i]+a[i+1]>=L && a[i+1]+a[i+2]<L) {
			num[i]=i+2;	num[i+1]=i+1;
			i=i+2;
		}
		else if (a[i]+a[i+1]<L && a[i+1]+a[i+2]>=L) {
			num[i]=i+1;	num[i+1]=i+2;
			i=i+2;
		}
	}
	
	if (judge==-1) 
		printf("Impossible\n");
	else {
		printf("Possible\n");
		for (i=0;i<n-1;i++) 
			printf("%d\n", num[i]);
	}
	
	return 0;
	
}