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 C (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <stdlib.h>

int v[100000];

int main() {
    FILE *fin, *fout;
    int n, k, i, maxim, a, b;
    fin = fopen( "knotpuzzle.in", "r" );
    fout = fopen( "knotpuzzle.out", "w" );
    fscanf( fin, "%d%d%d", &n, &k, &v[0] );
    maxim = 0;
    for( i = 1; i < n; i++ ) {
        fscanf( fin, "%d", &v[i] );
        if( v[i - 1] + v[i] > maxim ) {
            maxim = v[i - 1] + v[i];
            a = i - 1;
            b = i;
        }
    }
    if( maxim >= k ) {
        fprintf( fout, "Possible\n" );
        for( i = 1; i <= a; i++ ) {
            fprintf( fout, "%d\n", i );
        }
        for( i = n - 1; i > b; i-- ) {
            fprintf( fout, "%d\n", i );
        }
        fprintf( fout, "%d", b );
    }
    else {
        fprintf( fout, "Impossible" );
    }
    fclose( fin );
    fclose( fout );
    return 0;
}