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: Accepted)

#include <cstdio>
#include <vector>
#include <algorithm>
#include <set>
using namespace std;
#define sz(v) ((int)((v).size()))
#define all(v) (v).begin(), (v).end()
#define pb push_back
typedef long long ll;
void read(int& x){ scanf("%d",&x); }
template<typename T,typename... Args>
void read(T&a,Args&...b){ read(a); read(b...); }

struct t {
	ll val;
	int l, r;
	bool operator<(const t& oth) const {
		return l < oth.l;
	}
};

set<t> s;
int l;

vector<int> ans;

void add(t x){
	auto it = s.insert(x).first;
	if(it != s.begin()){
		auto il = it;
		--il;
		if(il->val + it->val >= l){
			ans.pb(il->r);
			t n = {il->val+it->val, il->l, it->r};
			s.erase(il); s.erase(it);
			add(n);
			return;
		}
	}
	auto ir = it; ++ir;
	if(ir != s.end()){
		if(it->val + ir->val >= l){
			ans.pb(it->r);
			t n = {it->val+ir->val, it->l, ir->r};
			s.erase(it); s.erase(ir);
			add(n);
			return;
		}
	}
}

int main(){
	int n;
	read(n, l);
	for(int i=1; i<=n; ++i){
		int a; read(a);
		add({a, i, i});
	}
	if(s.size() >= 2u){
		puts("Impossible");
	} else {
		puts("Possible");
		reverse(all(ans));
		for(int A:ans) printf("%d\n",A);
	}
	return 0;
}