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: Accepted)

use std::io;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;
use std::f64::consts::PI;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}


fn main() {
    let line: Vec<usize> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();
    let n = line[0];
    let l = line[1] as u64;

    let a: Vec<u64> = read_line().split_whitespace().map(|s| s.parse().unwrap()).collect();

    let mut pos = None;
    for i in 0..n-1 {
        if a[i] + a[i+1] >= l {
            pos = Some(i);
            break;
        }
    }

    if let Some(p) = pos {
        println!("Possible");
        for i in 0..p {
            println!("{}", i+1);
        }
        for i in (p+1..n-1).rev() {
            println!("{}", i+1);
        }
        println!("{}", p+1);

    } else {
        println!("Impossible");
    }
}