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 <math.h>
#include <string.h>
#include <stdlib.h>
int main(void)
{
	//変数の宣言
	int n,l;
	int a[100010];

	//よく使う変数
	int i,j,k;
	int flag=0;
	long int ans=0;
	int count=0;
	int temp,temp1,temp2;
	int max,min;
	int len;
	//データの読み込み
	
	scanf("%d %d",&n,&l);
	
	for(i=0;i<n;i++){
		scanf("%d",&a[i]);
	}
	
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	for(i=0;i<n-1;i++){
		if(a[i]+a[i+1]>=l){
			flag=1;
			break;
		}
	}
	
	if(flag==0){
		printf("Impossible");
	}else{
		printf("Possible\n");
		temp=i;
		for(i=1;i<=temp;i++){
			printf("%d\n",i);
		}
		for(i=n-1;i>temp;i--){
			printf("%d\n",i);
		}
	}
	
	
//	printf("計算部分終了\n");
	//出力
	
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}