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 <bits/stdc++.h>
using namespace std;

const int MAX_N = 100005;
int N;
int L;

int A[MAX_N];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> N >> L;

    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }
    
    int i = 0;
    int j = N-1;
    int l = 0;

    for (int i = 0; i < N; ++i) {
        l += A[i];
    }

    vector<int> t;
    t.clear();

    while (j > i) {
        if (l >= L) {
            if (A[i] > A[j]) {
                //cout << A[j] << endl;
                l -= A[j];
                t.push_back(j-1);
                j--;
            } else {
                //cout << A[i] << endl;
                l -= A[i];
                t.push_back(i);
                i++;
            }
        } else {
            cout << "Impossible" << endl;
            return 0;
        }
    }

    cout << "Possible" << endl;

    for (size_t i = 0; i < t.size(); ++i) {
        cout << t[i]+1 << endl;
    }

    return 0;
}