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 CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define IT iterator
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
#define debug puts("wzpakking")
#define y1 ysghysgsygsh
using namespace std;
#define vi vector<int>
int n,flg[500];
char s[500];
vi q;
vector<pii> ans;
vi get(vi x){
	if (x.size()==1) return x;
	int t=x.back();
	x.pop_back();
	vi v=get(x),res;
	res.PB(t);
	for (auto i:v) res.PB(i);
	res.PB(-t); 
	reverse(v.begin(),v.end());
	for (auto i:v) res.PB(-i);
	return res;
}
void walk(int &x,int y){
	for (;x<y;ans.PB(pii(++x,0)));
	for (;x>y;ans.PB(pii(--x,0)));
}
int main(){
	scanf("%d%s",&n,s);
	For(i,0,(1<<n)-1) s[i]-='0';
	For(i,0,(1<<n)-1) For(j,0,(1<<n)-1)
		if ((i&j)==j&&s[i]&&!s[j])
			return puts("Impossible"),0;
	For(i,0,(1<<n)-1) if (!s[i])
		For(j,0,(1<<n)-1) if (j!=i&&(i&j)==j&&!s[j]) flg[i]=1;
	For(i,0,(1<<n)-1) if (!s[i]&&!flg[i]){
		//cout<<i<<endl;
		vector<int> tmp;
		For(j,0,n-1) if (i&(1<<j)) tmp.PB(j+1);
		tmp=get(tmp);
		for (auto j:tmp) q.PB(j);
	}
	int x=0;
	ans.PB(pii(0,0));
	for (auto i:q){
		int st,ed;
		if (i>0) st=i-1,ed=i;
		else st=-i,ed=-i-1;
		walk(x,st);
		ans.PB(pii(st,1));
		ans.PB(pii(ed,1));
		ans.PB(pii(ed,0));
		x=ed;
	}
	walk(x,0);
	printf("Possible\n%d\n",ans.size()-1);
	for (auto i:ans) printf("%d %d\n",i.fi,i.se);
}