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;
use std::collections::VecDeque;

fn main() {
    let mut s = String::new();
    io::stdin().read_line(&mut s);
    let nl: Vec<u64> = s.trim().split(' ').map(|x| x.parse().unwrap()).collect();

    s = String::new();
    io::stdin().read_line(&mut s);
    let mut xs: VecDeque<u64> = VecDeque::new();
    let mut sval = 0u64;

    for n in s.trim().split(' ').map(|x| x.parse().unwrap()) {
        sval += n;
        xs.push_back(n);
    }

    let mut rs: Vec<u64> = Vec::new();
    let mut fidx = 1u64;
    let mut bidx = xs.len() as u64 - 1;
    while !xs.is_empty() {
        if xs.len() == 1 {
            break;
        }
        else if sval < nl[1] {
            rs.clear();
            break;
        }
        let &f = xs.front().unwrap();
        let &b = xs.back().unwrap();
        let minval = if b < f {
            rs.push(bidx);
            bidx -= 1;
            xs.pop_back();
            b
        } else {
            rs.push(fidx);
            fidx += 1;
            xs.pop_front();
            f
        };
        sval -= minval;

    }

    println!("{}", if rs.is_empty() {
        "Impossible"
    }
    else {
        "Possible"
    });

    for r in rs {
        println!("{}", r);
    }
}