Problem: There are N oases on a number line. The coordinate of the i-th oases from the left is x_i.

Camel hopes to visit all these oases. Initially, the volume of the hump on his back is V. When the volume of the hump is v, water of volume at most v can be stored. Water is only supplied at oases. He can get as much water as he can store at a oasis, and the same oasis can be used any number of times.

Camel can travel on the line by either walking or jumping:

* Walking over a distance of d costs water of volume d from the hump. A walk that leads to a negative amount of stored water cannot be done.
* Let v be the amount of water stored at the moment. When v>0, Camel can jump to any point on the line of his choice. After this move, the volume of the hump becomes v/2 (rounded down to the nearest integer), and the amount of stored water becomes 0.



For each of the oases, determine whether it is possible to start from that oasis and visit all the oases.

Constraints

* 2 ≤ N,V ≤ 2 × 10^5
* -10^9 ≤ x_1 < x_2 < ... < x_N ≤ 10^9
* V and x_i are all integers.

Input

Input is given from Standard Input in the following format:


N V
x_1 x_2 ... x_{N}


Output

Print N lines. The i-th line should contain `Possible` if it is possible to start from the i-th oasis and visit all the oases, and `Impossible` otherwise.

Examples

Input

3 2
1 3 6


Output

Possible
Possible
Possible


Input

7 2
-10 -4 -2 0 2 4 10


Output

Impossible
Possible
Possible
Possible
Possible
Possible
Impossible


Input

16 19
-49 -48 -33 -30 -21 -14 0 15 19 23 44 52 80 81 82 84


Output

Possible
Possible
Possible
Possible
Possible
Possible
Possible
Possible
Possible
Possible
Possible
Possible
Impossible
Impossible
Impossible
Impossible

Solution (in CPP):

Solution:
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std ;

#define l first
#define r second

const int M = 40 ;
const int N = 1200010 ;
const int MAX = 1001000001 ;

int dep ;
int maxn ;
int ans[N] ;
int n, m, k ;
int base[N] ;
int L[M][N] ;
int R[M][N] ;
int farfrl[N] ;
int farfrr[N] ;
pair<int, int> rg[N] ;

void Init_dp(){
    for (int j = m ; j >= 0 ; j >>= 1){
        R[++ dep][n + 1] = n + 1 ;
        for (int i = 1 ; i <= n ; ++ i)
            L[dep][i] = (base[i] - base[i - 1] > j) ? i : L[dep][i - 1] ;
        for (int i = n ; i >= 1 ; -- i)
            R[dep][i] = (base[i + 1] - base[i] > j) ? i : R[dep][i + 1] ;
        if (!j) break ;
    }
    return ;
}
void Auxiliary_dp(){
    for (int i = 0 ; i <= maxn ; i += 2)
        for (int j = 1 ; j <= dep ; ++ j)
            if (1 << (j - 1) & i){
                farfrl[i] = max(farfrl[i], R[j][ farfrl[i ^ (1 << (j - 1))] + 1]) ;
                farfrr[i] = min(farfrr[i], L[j][ farfrr[i ^ (1 << (j - 1))] - 1]) ;
            }
    return ;
}
int main(){
    cin >> n >> m ;
    base[0] = -MAX ;
    base[n + 1] = MAX ; k = -1 ;
    for (int i = 1 ; i <= n ; ++ i)
        scanf("%d", &base[i]) ; Init_dp() ;
    for (int i = 1 ; i <= n ; ++ i)
        rg[i].l = L[1][i], rg[i].r = R[1][i] ;
    sort(rg + 1, rg + n + 1) ; maxn = (1 << dep) - 1 ;
    for (int i = 0 ; i <= maxn ; ++ i) farfrr[i] = n + 1 ;
    k += unique(rg + 1, rg + n + 1) - rg ; Auxiliary_dp() ;
    for (int i = 1 ; i <= k ; ++ i)
        for (int j = 0 ; j <= maxn ; j += 2)
            if (farfrl[j] + 1 >= rg[i].l && farfrr[(maxn ^ j) ^ 1] - 1 <= rg[i].r)
                { for (int o = rg[i].l ; o <= rg[i].r ; ++ o) ans[o] = 1 ; break ; }
    for (int i = 1 ; i <= n ; ++ i) puts(ans[i] ? "Possible" : "Impossible") ; return 0 ;
}
/*
0 8
0 8
1 7
0 8
1 7
0 8
2 6
0 8
*/