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 REP(i,n) for (int i = 0; i < (n); i++)
#define RREP(i, s, n) for (int i = s; i < (n); i++)
#define ALL(a) a.begin(), a.end()
#define RALL(a) a.rbegin(), a.rend()
using namespace std;
using ll = long long;
typedef pair<int, int> pint;
typedef pair<ll, ll> pll;
const ll MOD = 1000000007;
const ll INF = MOD * MOD;
const int inf = (1<<29);

int main() {

// editorial AC
    int N, L;
    cin >> N >> L;

    vector<int> a(N);
    REP(i, N) {
        cin >> a[i];
    }

    int tmp = inf;
    for(int i = 0; i < N-1; i++) {
        if (a[i] + a[i+1] >= L) {
            tmp = i;
            break;
        }
    }

    if (tmp == inf) {
        cout << "Impossible" << endl;
        return 0;
    } else {
        cout << "Possible" << endl;
    }

    for (int i = 0; i < tmp; i++) {
        cout << i+1 << " ";
    }
    for (int i = N-2; i > tmp; i--) {
        cout << i+1 << " ";
    }

    cout << tmp+1 << endl;


// WA
    // ll N, L;
    // cin >> N >> L;
    
    // vector<pll> a;
    // ll sum = 0;
    // for(ll i; i < N; i++) {
    //     ll t;
    //     cin >> t;
    //     a.push_back(make_pair(t, i+1));
    //     sum += t;
    // }

    // vector<ll> ans; // ほどく順番

    // while(a.size() >= 2) {
    //     if (sum < L) {
    //         cout << "Impossible" << endl;
    //         return 0;
    //     }
    //     if (a[0].first < a[a.size()-1].first) {
    //         sum -= a[0].first;
    //         ans.push_back(a[0].second);
    //         a.erase(a.begin());
    //     } else {
    //         sum -= a[a.size()-1].first;
    //         ans.push_back(a[a.size()-1].second);
    //         a.erase(a.end()-1);
    //     }
    // }

    // cout << "Possible" << endl;
    // for (ll i = 0; i < N-1; i++) {
    //     cout << ans[i] << " ";
    // }

    // cout << endl;
    

    return 0;
}