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::io::*;

fn main() {
    let mut s: String = String::new();
    stdin().read_to_string(&mut s).ok();
    let mut itr = s.trim().split_whitespace();
    let n: usize = itr.next().unwrap().parse().unwrap();
    let l: usize = itr.next().unwrap().parse().unwrap();
    let a: Vec<usize> = (0..n)
        .map(|_| itr.next().unwrap().parse().unwrap())
        .collect();
    let mut saigo = n;
    for i in 0..n - 1 {
        if a[i] + a[i + 1] >= l {
            saigo = i;
        }
    }
    if saigo == n {
        println!("Impossible");
        return;
    }

    let mut out = Vec::new();
    writeln!(out, "Possible").ok();
    for i in 0..n - 1 {
        if i != saigo {
            writeln!(out, "{}", i + 1).ok();
        }
    }
    writeln!(out, "{}", saigo + 1).ok();
    stdout().write_all(&out).unwrap();
}