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)

#![allow(unused_imports, unused_variables, dead_code, non_snake_case, unused_macros)]
use std::io::{stdin, Read, StdinLock};
use std::str::FromStr;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;
fn getline() -> String {
    let mut res = String::new();
    std::io::stdin().read_line(&mut res).ok();
    res
}
macro_rules ! readl { ( $ t : ty ) => { { let s = getline ( ) ; s . trim ( ) . parse ::<$ t > ( ) . unwrap ( ) } } ; ( $ ( $ t : ty ) ,+ ) => { { let s = getline ( ) ; let mut iter = s . trim ( ) . split ( ' ' ) ; ( $ ( iter . next ( ) . unwrap ( ) . parse ::<$ t > ( ) . unwrap ( ) , ) * ) } } ; }
macro_rules ! readlvec { ( $ t : ty ) => { { let s = getline ( ) ; let iter = s . trim ( ) . split ( ' ' ) ; iter . map ( | x | x . parse ( ) . unwrap ( ) ) . collect ::< Vec <$ t >> ( ) } } }
macro_rules ! mvec { ( $ v : expr , $ s : expr ) => { vec ! [ $ v ; $ s ] } ; ( $ v : expr , $ s : expr , $ ( $ t : expr ) ,* ) => { vec ! [ mvec ! ( $ v , $ ( $ t ) ,* ) ; $ s ] } ; }
macro_rules ! debug { ( $ x : expr ) => { println ! ( "{}: {:?}" , stringify ! ( $ x ) , $ x ) } }
fn printiter<'a, T>(v: &'a T)
where
    &'a T: std::iter::IntoIterator,
    <&'a T as std::iter::IntoIterator>::Item: std::fmt::Display,
{
    for (i, e) in v.into_iter().enumerate() {
        if i != 0 {
            print!(" ");
        }
        print!("{}", e);
    }
    println!("");
}
struct ContestPrinter {
    s: String,
}
impl ContestPrinter {
    fn new() -> ContestPrinter {
        ContestPrinter { s: String::new() }
    }
    fn print<T>(&mut self, x: T)
    where
        T: std::fmt::Display,
    {
        self.s.push_str(format!("{}", x).as_str());
    }
    fn println<T>(&mut self, x: T)
    where
        T: std::fmt::Display,
    {
        self.s.push_str(format!("{}\n", x).as_str());
    }
}
impl std::ops::Drop for ContestPrinter {
    fn drop(&mut self) {
        print!("{}", self.s);
    }
}
static MOD: i64 = 1e9 as i64 + 7;
fn is_max_i64(num: i64) -> bool {
    if num == i64::max_value() {
        true
    } else {
        false
    }
}
fn main() {
    let mut pr = ContestPrinter::new();
    let (N, L) = readl!(usize, u64);
    let v = readlvec!(u64);
    let mut sum: u64 = v.iter().sum();

    let mut l = 0;
    let mut r = N-1;
    let mut out = Vec::new();
    loop {
        if sum < L {
            pr.println("Impossible");
            return;
        }

        let lv = v[l];
        let rv = v[r];
        if (l + 1) == r {
            out.push(r);
            break;
        }
        if lv < rv {
            sum -= lv;
            l += 1;
            out.push(l);
        } else {
            sum -= rv;
            out.push(r);
            r -= 1;
        }
    }

    pr.println("Possible");
    for o in out { pr.println(o); }
}