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 <iostream>
#include <queue>
using namespace std;

int a[1000000005];
queue<int> q;

int main(void) {
	int N, L;
	cin.tie(0);
	ios::sync_with_stdio(false);

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


	int l = 0, r = N-1, ans = 0, k = 0;
	long long int sum = 0;
	for (int i = 0; i < N - 1; i++) {
		sum = 0;
		for (int j = l; j <= r; j++) {
			sum += a[j];
			if (sum >= L) {
				if (a[l] <= a[r]) {
					a[l] = 0;
					q.push(l+1);
					l++;
				} else if (a[l] > a[r]) {
					a[r] = 0;
					q.push(r);
					r--;
				}
				break;
			}
		}
		if (l == r) {
			ans = 1;
			break;
		}

	}

	if (ans) {
		cout << "Possible\n";
		for (int i = 0; i < N - 1; i++) {
			cout << q.front() << endl;
			q.pop();
		}
	} else {
		cout << "Impossible\n";
	}


	return 0;
}