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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include<stdio.h>
typedef long long ll;
int N,L;
ll s[100002];
int ans[100002];
int main(){
  int i;
  int a;
  scanf("%d %d",&N,&L);
    scanf("%d",&a);
    s[1] = a;
  for(i = 2; i <= N; i++){
    scanf("%d",&a);
    s[i] = s[i-1]+a;    
  }
 
  if(s[N] < L){printf("Impossible\n");return 0;}
  
  int l = 1, r = N,j = 0;
  while(r != l){
    if(s[r]-s[l-1] < L)break;
    
    if(s[r]-s[r-1] > s[l]-s[l-1]){
      ans[j++] = l++;
    }
    else {ans[j++] = r-1;
      r--;
    }
  }
  if(r == l){
    printf("Possible\n");
    for(i = 0; i < j; i++){
      printf("%d\n",ans[i]);
    }
  }
  else printf("Impossible\n");
  return 0;
}