#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(int i=0;i<(int)(n);i++)
#define N 200010

vector<int>b[8];
vector<int>ax;
vector<int>ay;


int main() {
	char a[N];
	f(i, N)a[i] = 0;
	int c[10];
	int n, k;
	int x, y, z;
	int s, ans;
	bool v = true;
	ans = 0;
	b[0].push_back(1);
	f(i, 7) {
		f(j, b[i].size()) {
			b[i + 1].push_back(b[i][j]);
		}
		b[i + 1].push_back(i + 2);
		for(int j=b[i].size()-1;j>=0;j--){
			b[i + 1].push_back(-b[i][j]);
		}
		b[i + 1].push_back(-(i + 2));
	}
	scanf("%d", &n);
	scanf("%s", a);
	k = strlen(a);
	if (a[0] == '0') {
		printf("Impossible\n");
		return 0;
	}
	x = 1;
	ax.push_back(0);
	ay.push_back(0);
	f(i, k) {
		f(j, i) {
			if ((i&j) == j) {
				if ((a[i] == '1') && (a[j] == '0')) {
					printf("Impossible\n");
					return 0;
				}
			}
		}
	}
	f(i, n) {
		f(j, x) {
			if (a[j] == '1') {
				if (a[x + j] == '0') {
					y = 1;
					z = x + j;
					f(ii, n) {
						if (z % 2 == 1) {
							c[y] = ii + 1;
							y++;
						}
						z /= 2;
					}
					y -= 2;
					f(ii, b[y].size()) {
						if (b[y][ii] > 0) {
							for (int j = 1; j <= c[b[y][ii]] - 1; j++) {
								ax.push_back(j);
								ay.push_back(0);
							}
							ax.push_back(c[b[y][ii]]);
							ay.push_back(0);
							ax.push_back(c[b[y][ii]]);
							ay.push_back(1);
							ax.push_back(c[b[y][ii]]-1);
							ay.push_back(1);
							for (int j = c[b[y][ii]] - 1; j >= 0; j--) {
								ax.push_back(j);
								ay.push_back(0);
							}
						}
						else {
							for (int j = 1; j <= c[-b[y][ii]] - 1; j++) {
								ax.push_back(j);
								ay.push_back(0);
							}
							ax.push_back(c[-b[y][ii]] - 1);
							ay.push_back(1);
							ax.push_back(c[-b[y][ii]]);
							ay.push_back(1);
							ax.push_back(c[-b[y][ii]]);
							ay.push_back(0);
							for (int j = c[-b[y][ii]] - 1; j >= 0; j--) {
								ax.push_back(j);
								ay.push_back(0);
							}
						}
					}
				}
			}
			else {
				if (a[x + j] == '1') {
					printf("Impossible\n");
					return 0;
				}
			}
		}

		x *= 2;
	}
	printf("Possible\n");
	printf("%d\n", ax.size() - 1);
	f(i, ax.size()) {
		printf("%d %d\n", ax[i], ay[i]);
	}


	return 0;
}
### Prompt

Generate a CPP solution to the following 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

### Response

```cpp
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<functional>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
using namespace std;
#define MOD 1000000007
#define f(i,n) for(int i=0;i<(int)(n);i++)
#define N 200010

vector<int>b[8];
vector<int>ax;
vector<int>ay;


int main() {
	char a[N];
	f(i, N)a[i] = 0;
	int c[10];
	int n, k;
	int x, y, z;
	int s, ans;
	bool v = true;
	ans = 0;
	b[0].push_back(1);
	f(i, 7) {
		f(j, b[i].size()) {
			b[i + 1].push_back(b[i][j]);
		}
		b[i + 1].push_back(i + 2);
		for(int j=b[i].size()-1;j>=0;j--){
			b[i + 1].push_back(-b[i][j]);
		}
		b[i + 1].push_back(-(i + 2));
	}
	scanf("%d", &n);
	scanf("%s", a);
	k = strlen(a);
	if (a[0] == '0') {
		printf("Impossible\n");
		return 0;
	}
	x = 1;
	ax.push_back(0);
	ay.push_back(0);
	f(i, k) {
		f(j, i) {
			if ((i&j) == j) {
				if ((a[i] == '1') && (a[j] == '0')) {
					printf("Impossible\n");
					return 0;
				}
			}
		}
	}
	f(i, n) {
		f(j, x) {
			if (a[j] == '1') {
				if (a[x + j] == '0') {
					y = 1;
					z = x + j;
					f(ii, n) {
						if (z % 2 == 1) {
							c[y] = ii + 1;
							y++;
						}
						z /= 2;
					}
					y -= 2;
					f(ii, b[y].size()) {
						if (b[y][ii] > 0) {
							for (int j = 1; j <= c[b[y][ii]] - 1; j++) {
								ax.push_back(j);
								ay.push_back(0);
							}
							ax.push_back(c[b[y][ii]]);
							ay.push_back(0);
							ax.push_back(c[b[y][ii]]);
							ay.push_back(1);
							ax.push_back(c[b[y][ii]]-1);
							ay.push_back(1);
							for (int j = c[b[y][ii]] - 1; j >= 0; j--) {
								ax.push_back(j);
								ay.push_back(0);
							}
						}
						else {
							for (int j = 1; j <= c[-b[y][ii]] - 1; j++) {
								ax.push_back(j);
								ay.push_back(0);
							}
							ax.push_back(c[-b[y][ii]] - 1);
							ay.push_back(1);
							ax.push_back(c[-b[y][ii]]);
							ay.push_back(1);
							ax.push_back(c[-b[y][ii]]);
							ay.push_back(0);
							for (int j = c[-b[y][ii]] - 1; j >= 0; j--) {
								ax.push_back(j);
								ay.push_back(0);
							}
						}
					}
				}
			}
			else {
				if (a[x + j] == '1') {
					printf("Impossible\n");
					return 0;
				}
			}
		}

		x *= 2;
	}
	printf("Possible\n");
	printf("%d\n", ax.size() - 1);
	f(i, ax.size()) {
		printf("%d %d\n", ax[i], ay[i]);
	}


	return 0;
}
```