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>
using namespace std;

#define Fi first
#define Se second
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define szz(x) (int)(x).size()
#define rep(i, n) for(int i=0;i<(n);i++)
typedef tuple<int, int, int> t3;

int N;
char buf[330];
int A[330];

vector <int> Get(vector <int> X) {
	if(szz(X) == 1) return {X[0]};
	else {
		int t = X.back();
		X.pop_back();
		auto w = Get(X);
		vector <int> res;
		res.pb(t);
		res.insert(res.end(), all(w));
		res.pb(-t);
		reverse(all(w));
		for(int e : w) res.pb(-e);
		return res;
	}
}

int main() {
	scanf("%d", &N);
	scanf("%s", buf);
	for(int i=0;i<1<<N;i++) A[i] = buf[i] - '0';
	int B = 1<<N;
	rep(i, B) rep(j, B) if((i&j) == j) {
		if(A[i] && A[j] == 0) {
			puts("Impossible");
			return 0;
		}
	}
	vector <int> w;
	rep(i, B) if(A[i] == 0) {
		int ok = 1;
		rep(j, i) if((i&j) == j && A[j] == 0) ok = 0;
		if(ok) w.pb(i);
	}
	vector <int> R;
	for(int e : w) {
		vector <int> t;
		rep(i, N) if(1<<i & e) t.pb(i + 1);
		auto v2 = Get(t);
		R.insert(R.end(), all(v2));
	}
	vector <pii> ans;
	int now = 0;
	ans.pb({0, 0});
	auto Move = [&](int x) {
		while(now < x) ans.pb({++now, 0});
		while(now > x) ans.pb({--now, 0});
	};
	for(int e : R) {
		int st, en;
		if(e > 0) st = e - 1, en = e;
		else st = -e, en = -e - 1;
		Move(st);
		ans.pb({st, 1});
		ans.pb({en, 1});
		ans.pb({en, 0});
		now = en;
	}
	Move(0);
	puts("Possible");
	printf("%d\n", szz(ans) - 1);
	for(pii e : ans) printf("%d %d\n", e.Fi, e.Se);
	return 0;
}