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: Runtime Error)

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <sstream>
#include <typeinfo>
#include <fstream>

#define DIV 1000000007

using namespace std;

long long N, L;
long long a[100005];

int main(){
	cin >> N >> L;
	long long maxcomb = -1;
	long long maxidx = -1;

	for(int i = 0; i < N; i++){
		cin >> a[i];
	}

	for(int i = 0; i < N-1; i++){
		if(maxcomb < a[i] + a[i+1]){
			maxidx = i;
			maxcomb = a[i] + a[i+1];
		}
	}
	if(maxcomb < L){
		cout << "Impossible" << endl;
		return 0;
	}

	cout << "Possible" << endl;
	for(int i = 0; i < N - 1; i++){
		return 1;
		if(i != maxidx){
			cout << i+1 << endl;
		}
	}
	cout << maxidx+1 << endl;




}