Problem: Given are a positive integer N and a sequence of length 2^N consisting of 0s and 1s: A_0,A_1,\ldots,A_{2^N-1}. Determine whether there exists a closed curve C that satisfies the condition below for all 2^N sets S \subseteq \\{0,1,\ldots,N-1 \\}. If the answer is yes, construct one such closed curve.

* Let x = \sum_{i \in S} 2^i and B_S be the set of points \\{ (i+0.5,0.5) | i \in S \\}.
* If there is a way to continuously move the closed curve C without touching B_S so that every point on the closed curve has a negative y-coordinate, A_x = 1.
* If there is no such way, A_x = 0.



For instruction on printing a closed curve, see Output below.

Constraints

* 1 \leq N \leq 8
* A_i = 0,1 \quad (0 \leq i \leq 2^N-1)
* A_0 = 1

Input

Input is given from Standard Input in the following format:


N
A_0A_1 \cdots A_{2^N-1}


Output

If there is no closed curve that satisfies the condition, print `Impossible`.

If such a closed curve exists, print `Possible` in the first line. Then, print one such curve in the following format:


L
x_0 y_0
x_1 y_1
:
x_L y_L


This represents the closed polyline that passes (x_0,y_0),(x_1,y_1),\ldots,(x_L,y_L) in this order.

Here, all of the following must be satisfied:

* 0 \leq x_i \leq N, 0 \leq y_i \leq 1, and x_i, y_i are integers. (0 \leq i \leq L)
* |x_i-x_{i+1}| + |y_i-y_{i+1}| = 1. (0 \leq i \leq L-1)
* (x_0,y_0) = (x_L,y_L).



Additionally, the length of the closed curve L must satisfy 0 \leq L \leq 250000.

It can be proved that, if there is a closed curve that satisfies the condition in Problem Statement, there is also a closed curve that can be expressed in this format.

Examples

Input

1
10


Output

Possible
4
0 0
0 1
1 1
1 0
0 0


Input

2
1000


Output

Possible
6
1 0
2 0
2 1
1 1
0 1
0 0
1 0


Input

2
1001


Output

Impossible


Input

1
11


Output

Possible
0
1 1

Solution (in PYTHON):

Solution:
def rev(a):
    return [-v for v in a[::-1]]

def make(s):
    assert len(s)>0
    pre_s = s[:-1]
    v = s[-1]
    if len(pre_s)==0:
        return [v]
    return make(pre_s) + [v] + rev(make(pre_s)) + [-v]

def make_pre_path(v):
    if v>0:
        return [(i,0) for i in range(1,v+1)] + [(v,1), (v-1,1)] + [(i,0) for i in range(v-1,0,-1)]
    elif v<0:
        return make_pre_path(-v)[::-1]
    assert False

def make_path(v):
    return make_pre_path(v)+[(0,0)]

def test():
    #print(make([1,2,3]))
    main("3","11111110")
    #main("8","1"+"0"*(2**8-1))

def main(s1,s2):
    ok = True
    n = int(s1)
    ret_gp = []
    for i in range(2**n-1,-1,-1):
        s = [v+1 for v in range(n) if (i>>v)%2==1 ]
        if s2[i]=="0":
            for ind in range(n):
                if (i>>ind)%2==0 and s2[(i+2**ind)]=="1":
                    ok = False
            if len(s)>0:
                ret_gp += make(s)
    ret_path = [(0,0)] + [p for v in ret_gp for p in make_path(v)]
    if 0:
        print(ret_gp)
    if ok:
        print("Possible")
        print(len(ret_path) - 1)
        for x,y in ret_path:
            print(str(x)+" "+str(y))
    else:
        print("Impossible")

#test()
        
s1 = raw_input()
s2 = raw_input()
main(s1,s2)