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>
using namespace std;

#define REPU(i, a, b) for (int i = (a); i < (b); ++i)
#define REPD(i, a, b) for (int i = (a); i > (b); --i)
#define MEM(a, x) memset(a, x, sizeof(a))
#define ALL(a) a.begin(), a.end()
#define UNIQUE(a) a.erase(unique(ALL(a)), a.end())
#define DEBUG(x) cerr << #x << " = " << x << endl

typedef long long ll;
const int MOD = 1000000007;

template<class T, class U> inline T tmin(T a, U b) { return (a < b) ? a : b; }
template<class T, class U> inline T tmax(T a, U b) { return (a > b) ? a : b; }
template<class T, class U> inline void amax(T &a, U b) { if (b > a) a = b; }
template<class T, class U> inline void amin(T &a, U b) { if (b < a) a = b; }
template<class T> inline T tabs(T a) { return (a > 0) ? a : -a; }
template<class T> T gcd(T a, T b) { while (b != 0) { T c = a; a = b; b = c % b; } return a; }

const int N = 100005;
int a[N];

int main(int argc, char *argv[]) {
	ios_base::sync_with_stdio(false);
	
	int n, L;

	cin >> n >> L;
	REPU(i, 0, n) {
		cin >> a[i];
	}
	int ind = -1;
	REPU(i, 1, n) {
		if (a[i] + a[i - 1] >= L) {
			ind = i; break;
		}
	}
	if (ind == -1) {
		puts("Impossible"); return 0;
	}
	puts("Possible");
	REPU(i, 1, ind) {
		printf("%d\n", i);
	}
	REPD(i, n - 1, ind) {
		printf("%d\n", i);
	}
	printf("%d\n", ind);
	
	return 0;
}