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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int,int> P;
int INF = 1e9+7;
int mod = 998244353;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
signed main(){
    int N,L;
    cin >> N >> L;
    vector<int>a(N);
    vector<int>ans;
    int sum = 0;
    for(int i = 0; i < N; i++) {
        cin >> a[i];
        sum += a[i];
    }
    int now = 0,now2 = N-1;
    while(now < now2) {
        if(sum < L) {
            cout << "Impossible" << endl;
            return 0;
        }
        int x = a[now],y = a[now2];
        if(x > y) {
            ans.push_back(now2);
            sum -= y;
            now2--;
        }
        else {
            ans.push_back(now+1);
            sum -= x;
            now++;
        }
    }
    cout << "Possible" << endl;
    for(int x:ans) {
        cout << x << endl;
    }
}