# include <bits/stdc++.h>
using namespace std;
namespace Base{
	# define mr make_pair
	typedef long long ll;
	typedef double db;
	const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
	const ll  infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
	template<typename T> void read(T &x){
    	x = 0; int fh = 1; double num = 1.0; char ch = getchar();
		while (!isdigit(ch)){ if (ch == '-') fh = -1; ch = getchar(); }
		while (isdigit(ch)){ x = x * 10 + ch - '0'; ch = getchar(); }
	    if (ch == '.'){
	    	ch = getchar();
	    	while (isdigit(ch)){num /= 10; x = x + num * (ch - '0'); ch = getchar();}
		}
		x = x * fh;
	}
	template<typename T> void chmax(T &x, T y){x = x < y ? y : x;}
	template<typename T> void chmin(T &x, T y){x = x > y ? y : x;}
}
using namespace Base;

const int N = 300010;
int dx, dy, total, qx[N], qy[N], n, use[N], nd[N], cnt, now[N], ths[N], nxt[N];
char s[N];
void moveto(int x){
	if (x > dx) for (int i = dx + 1; i <= x; i++) qx[++total] = i, qy[total] = dy;
		else for (int i = dx - 1; i >= x; i--) qx[++total] = i, qy[total] = dy;
	dx = x;
}

void moveup(){
	qx[++total] = dx, qy[total] = ++dy;
}
void movedown(){
	qx[++total] = dx, qy[total] = --dy;
}
int main(){
	read(n);
	scanf("\n%s", s);
	bool flag = true;
	for (int i = 0; i < (1 << n); i++)
		if (s[i] == '0'){
			if (use[i] == 0) nd[++cnt] = i; 
			for (int j = 0; j < (1 << n); j++)
				if ((i | j) == j){
					if (s[j] == '1') flag = false;
					use[j] = true;
				}
		}
	if (flag == false){
		puts("Impossible");
		return 0;
	}
	
	dx = 0, dy = 0;
	for (int i = 1; i <= cnt; i++){
		int num = 0, len = 2, nlen;
		for (int j = 1; j <= n; j++)
			if (nd[i] & (1 << (j - 1))) now[++num] = j;
		ths[1] = now[num]; ths[2] = -now[num];
		for (int j = num - 1; j >= 1; j--){
			nxt[nlen = 1] = now[j];
			for (int k = 1; k <= len; k++) nxt[++nlen] = ths[k];
			nxt[++nlen] = now[j];
			nxt[++nlen] = -now[j];
			for (int k = len; k >= 1; k--) nxt[++nlen] = ths[k];
			nxt[++nlen] = -now[j];
			len = nlen;
			for (int k = 1; k <= len; k++) ths[k] = nxt[k];
		}
		for (int j = 1; j <= len; j++){
			int x = abs(ths[j]);
			if (dx < x){
				moveto(x - 1); 
				if (ths[j] > 0) moveup(), moveto(x), movedown();
					else moveto(x);
			}
			else{
				moveto(x);
				if (ths[j] > 0) moveup(), moveto(x - 1), movedown();
					else moveto(x - 1);
			}
		}
		moveto(0);
	}
	
	puts("Possible");
	printf("%d\n", total);
	for (int i = 0; i <= total; i++)
		printf("%d %d\n", qx[i], qy[i]);
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
# include <bits/stdc++.h>
using namespace std;
namespace Base{
	# define mr make_pair
	typedef long long ll;
	typedef double db;
	const int inf = 0x3f3f3f3f, INF = 0x7fffffff;
	const ll  infll = 0x3f3f3f3f3f3f3f3fll, INFll = 0x7fffffffffffffffll;
	template<typename T> void read(T &x){
    	x = 0; int fh = 1; double num = 1.0; char ch = getchar();
		while (!isdigit(ch)){ if (ch == '-') fh = -1; ch = getchar(); }
		while (isdigit(ch)){ x = x * 10 + ch - '0'; ch = getchar(); }
	    if (ch == '.'){
	    	ch = getchar();
	    	while (isdigit(ch)){num /= 10; x = x + num * (ch - '0'); ch = getchar();}
		}
		x = x * fh;
	}
	template<typename T> void chmax(T &x, T y){x = x < y ? y : x;}
	template<typename T> void chmin(T &x, T y){x = x > y ? y : x;}
}
using namespace Base;

const int N = 300010;
int dx, dy, total, qx[N], qy[N], n, use[N], nd[N], cnt, now[N], ths[N], nxt[N];
char s[N];
void moveto(int x){
	if (x > dx) for (int i = dx + 1; i <= x; i++) qx[++total] = i, qy[total] = dy;
		else for (int i = dx - 1; i >= x; i--) qx[++total] = i, qy[total] = dy;
	dx = x;
}

void moveup(){
	qx[++total] = dx, qy[total] = ++dy;
}
void movedown(){
	qx[++total] = dx, qy[total] = --dy;
}
int main(){
	read(n);
	scanf("\n%s", s);
	bool flag = true;
	for (int i = 0; i < (1 << n); i++)
		if (s[i] == '0'){
			if (use[i] == 0) nd[++cnt] = i; 
			for (int j = 0; j < (1 << n); j++)
				if ((i | j) == j){
					if (s[j] == '1') flag = false;
					use[j] = true;
				}
		}
	if (flag == false){
		puts("Impossible");
		return 0;
	}
	
	dx = 0, dy = 0;
	for (int i = 1; i <= cnt; i++){
		int num = 0, len = 2, nlen;
		for (int j = 1; j <= n; j++)
			if (nd[i] & (1 << (j - 1))) now[++num] = j;
		ths[1] = now[num]; ths[2] = -now[num];
		for (int j = num - 1; j >= 1; j--){
			nxt[nlen = 1] = now[j];
			for (int k = 1; k <= len; k++) nxt[++nlen] = ths[k];
			nxt[++nlen] = now[j];
			nxt[++nlen] = -now[j];
			for (int k = len; k >= 1; k--) nxt[++nlen] = ths[k];
			nxt[++nlen] = -now[j];
			len = nlen;
			for (int k = 1; k <= len; k++) ths[k] = nxt[k];
		}
		for (int j = 1; j <= len; j++){
			int x = abs(ths[j]);
			if (dx < x){
				moveto(x - 1); 
				if (ths[j] > 0) moveup(), moveto(x), movedown();
					else moveto(x);
			}
			else{
				moveto(x);
				if (ths[j] > 0) moveup(), moveto(x - 1), movedown();
					else moveto(x - 1);
			}
		}
		moveto(0);
	}
	
	puts("Possible");
	printf("%d\n", total);
	for (int i = 0; i <= total; i++)
		printf("%d %d\n", qx[i], qy[i]);
	return 0;
}
```