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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
 
#define INF 1000000000
 
#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))

int main() {
	int N, L;
	int a[100001];
	int i, j;
	int left, right;
	
	scanf("%d %d", &N, &L);
	for(i=1;i<=N;i++) {
		scanf("%d", &a[i]);
	}
	
	for(i=1;i<=N-1;i++) {
		if(a[i] + a[i+1] >= L) {
			left = i;
			break;
		}
	}
	if(i == N) {
		printf("Impossible\n");
		return 0;
	}
	
	
	printf("Possible\n");
	for(i=1;i<left;i++) {
		printf("%d\n", i);
	}
	for(i=N-1;i>=left;i--) {
		printf("%d\n", i);
	}
	
	/*
	left = 1;
	right = N;
	for(i=1;i<=N-1;i++) {
		if(a[left] < a[right]) {
			printf("%d\n", left++);
		} else {
			printf("%d\n", --right);
		}
	}
	*/
	
	return 0;
}