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 <bits/stdc++.h>
#define int long long
#define mod (int)(1e9+7)
#define inf (int)(3e18+7)
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define P pair<int,int>
#define PiP pair<int,pair<int,int>>
#define all(v) v.begin(),v.end()
#define mkp make_pair
#define mkt make_tuple
#define prique(T) priority_queue<T,vector<T>,greater<T>>
#define vecunique(vec) sort(vec.begin(), vec.end());decltype(vec)::iterator result = std::unique(vec.begin(), vec.end());vec.erase(result, vec.end())
using namespace std;

bool prime(int x) {
	for (int i = 2; i * i <= x; i++) {
		if (x % i == 0)return false;
	}
	return x > 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	return x / gcd(x, y) * y;
}
int kai(int x, int y) {
	int res = 1;
	for (int i = x - y + 1; i <= x; i++) {
		res *= i; res %= mod;
	}
	return res;
}
int mod_pow(int x, int y) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % mod;
		}
		x = x * x % mod;
		y >>= 1;
	}
	return res;
}
int comb(int x, int y) {
	if (y > x)return 0;
	return kai(x, y) * mod_pow(kai(y, y), mod - 2) % mod;
}
/*--------Library Zone!--------*/

int n, a[114514], l;
signed main() {
	cin >> n >> l;
	rep(i, n)cin >> a[i];
	int memo;
	rep(i, n - 1) {
		if (a[i] + a[i + 1] >= l) {
			memo = i;
			goto ok;
		}
	}
	cout << "Impossible" << endl;
	return 0;
ok:;
	cout << "Possible" << endl;
	rep(i, memo)cout << i + 1 << endl;
	for (int i = n - 2; i > memo; i--)cout << i + 1 << endl;
	cout << memo + 1 << endl;
}