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;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;

/*
 <url:https://agc002.contest.atcoder.jp/tasks/agc002_c>
問題文============================================================
 N 本のロープがあります。 ロープは 1 から N まで番号が振られています。 ロープ i の長さは ai です。
 
 最初、1≤i≤N−1 について、ロープ i の右端とロープ i+1 の左端が結ばれています。
 高橋君は次の操作を N−1 回行い、すべての結び目をほどこうとしています。
 
 ひと繋がりのロープのうち、長さの総和が L 以上のものをひとつ選ぶ。
 選んだひと繋がりのロープに結び目があれば、結び目のうちどれかひとつをほどく。
 
 高橋君は結び目をほどく順番を工夫し、すべての結び目をほどくことができるでしょうか？
 可能か判定し、可能ならば結び目をほどく順番をひとつ求めてください。

=================================================================

解説=============================================================

================================================================
*/

struct sec{
    ll l,r,len;
    sec(){}
    sec(ll l,ll r,ll len):l(l),r(r),len(len){}
};
void solve(){
    int N,L; cin >> N >> L;
    vector<int> a(N);
    for(auto &in:a) cin >> in;
    for(int i = 0; i < N-1;i++){
        if(a[i] + a[i+1] >= L){
            cout << "Possible" << endl;
            for(int j = 1; j <= i;j++) cout << j << endl;
            for(int j = N - 1; j > i; j--) cout << j << endl;
            return;
        }
    }
    cout << "Impossible" << endl;
}
int main(void) {
	cin.tie(0); ios::sync_with_stdio(false);
    solve();
	return 0;
}