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: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <bitset>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<pint> vpint;
#define mp make_pair
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for (int i=(n)-1;i>=0;i--)
#define reps(i,f,n) for(int i=(f);i<(n);i++)
#define REPS(i,f,n) for (int i=(f)-1;i>=(n);i--)
const int mod = 1e9;

int main(void){
	int n, l; cin >> n >> l;
	vector<ll> a;
	long long sum = 0;
	rep(i, n){
		int tmp;
		cin >> tmp;
		sum += tmp;
		a.push_back(tmp);
	}
	int L = 1, r = n - 1;
	vector<ll> ans;
 	rep(i, n - 1){
 		// printf("%lld %lld %lld\n", a[0], a[a.size() - 1], sum);
		if(a[0] < a[a.size() - 1] && sum >= l){
			sum -= a[0];
			a.erase(a.begin());
			ans.push_back(L);
			L++;
			continue;
		}else if(a[0] >= a[a.size() - 1] && sum >= l){
			sum -= a[a.size() - 1];
			a.pop_back();
			ans.push_back(r);
			r--;
			continue;
		}
		printf("Impossible\n");
		return 0;
	}
	printf("Possible\n");
	rep(i, n - 1){
		printf("%lld\n", ans[i]);
	}

	return 0;
}