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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::collections::VecDeque;

fn read1<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn readn<T: std::str::FromStr>(delimiter: &str) -> Vec<T> {
    let s = read1::<String>();
    s.split(delimiter).map(|e| e.parse().ok().unwrap()).collect::<Vec<T>>()
}

fn main() {
    let nl = readn::<usize>(" ");
    let n = nl[0];
    let l = nl[1];

    let mut a = readn::<usize>(" ");
    let mut k = a.iter().sum::<usize>();
    let mut que = VecDeque::from(a);

    let mut nf = 1;
    let mut nb = n - 1;
    let mut ans0 = "Possible";
    let mut ans1 = "".to_string();
    for _ in 1..n {
        if k < l {
            ans0 = "Impossible";
            break;
        }
        let f = *(que.front().unwrap());
        let b = *(que.back().unwrap());
        if f < b {
            ans1 = ans1 + &(nf.to_string()) + "\n";
            nf += 1;
            k -= f;
            que.pop_front();
        } else {
            ans1 = ans1 + &(nb.to_string()) + "\n";
            nb -= 1;
            k -= b;
            que.pop_back();
        }
    }

    if ans0 == "Possible" {
        println!("{}", ans0);
        println!("{}", ans1);
    } else {
        println!("{}", ans0);
    }
}