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;

vector<int> rec(int x) {
	vector <int> sol;
	int pet = 1, bit = 0;
	for (; bit < 10; bit++) {
		pet = 1<<bit;
		if ((pet&x) != 0) break;
	}
	if (pet == x) {
		sol.push_back(bit+1);
		return sol;
	}
	vector<int> aux = sol  = rec(x^pet);
	sol.push_back(bit+1);
	for (auto it = aux.rbegin(); it != aux.rend(); it++)
		sol.push_back(-*it);
	sol.push_back(-bit-1);
	return sol;
}

string tira(const vector<int> & par) {
	string sol;
	for (int x: par) {
		sol += string(abs(x)-1,'R');
		if (x > 0) sol += "URDL";
		else sol += "RULD";
		sol += string(abs(x)-1,'L');
	}
	return sol;
}


int main() {
  ios_base::sync_with_stdio(false); 
  cin.tie(nullptr); 

	int n;
	cin >> n;
	string s;
	cin >> s;
	int m = 1<<n;
	bool ok = true;
	vector<int> obs;
	for (int i = 1; i < m; i++) {
		if (s[i] == '0') obs.push_back(i);
		for (int j = i+1; j < m; j++) {
			int p = i, g = j;
			if (p > g) swap(p, g);
			if ((p&g) == p and s[p] == '0' and s[g] == '1') ok = false;
		}
	}
	if (!ok) cout << "Impossible" << endl;
	else {
		cout << "Possible" << endl;
		string sol;
		for (auto x: obs)  {
			sol += tira(rec(x));
		}
		int k = sol.size();
		cout << k   << endl;
		assert(k <= 250000);
		int px = 0, py = 0; 
		cout << px << " " << py << "\n";
		for (int i = 0; i < k; i++) {
			if (sol[i] == 'U') py++;
			else if (sol[i] == 'D') py--;
			else if (sol[i] == 'L') px--;
			else if (sol[i] == 'R') px++;
			cout << px << " " << py << "\n";
		}
	}


}