#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

double startTime;
double getCurrentTime() {
	return ((double)clock() - startTime) / CLOCKS_PER_SEC;
}

const int N = 277;
int n;
char s[N];
vector<pii> a[N];

int main()
{
	startTime = (double)clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d %s", &n, s);
//	n = 8;
//	for (int i = 0; i < (1 << n); i++)
//		s[i] = '1';
//	s[(1 << n) - 1] = '0';

	if (s[0] == '0') {
		printf("Impossible\n");
		return 0;
	}
	for (int i = 0; i < (1 << n); i++)
		for (int j = 0; j < (1 << n); j++) {
			if (s[i] == '0' && s[j] == '1' && ((i & j) == i)) {
				printf("Impossible\n");
				return 0;
			}
		}
	for (int i = 0; i < n; i++) {
		int mask = 1 << i;
		for (int j = 0; j <= i + 1; j++)
			a[mask].push_back(mp(j, 0));
		a[mask].push_back(mp(i + 1, 1));
		a[mask].push_back(mp(i, 1));
		for (int j = i; j >= 0; j--)
			a[mask].push_back(mp(j, 0));
	}
	for (int mask = 1; mask < (1 << n); mask++) {
		if (!a[mask].empty()) continue;
		int p = 0;
		while(((mask >> p) & 1) == 0) p++;
		p = 1 << p;
		int q = mask ^ p;
		a[mask].push_back(mp(0, 0));
		for (int i = 1; i < (int)a[p].size(); i++)
			a[mask].push_back(a[p][i]);
		for (int i = 1; i < (int)a[q].size(); i++)
			a[mask].push_back(a[q][i]);
		for (int i = (int)a[p].size() - 2; i >= 0; i--)
			a[mask].push_back(a[p][i]);
		for (int i = (int)a[q].size() - 2; i >= 0; i--)
			a[mask].push_back(a[q][i]);
	}
	vector<pii> ans;
	ans.push_back(mp(0, 0));
	for (int i = 0; i < (1 << n); i++) {
		if (s[i] == '1') continue;
		bool ok = true;
		for (int j = 0; j < i; j++) {
			if ((j & i) == j)
				ok &= s[j] == '1';
		}
		if (!ok) continue;
		for (int j = 1; j < (int)a[i].size(); j++)
			ans.push_back(a[i][j]);
	}
	printf("Possible\n");
	printf("%d\n", (int)ans.size() - 1);
	for (pii t : ans)
		printf("%d %d\n", t.first, t.second);

	return 0;
}

### Prompt

In cpp, your task is to solve 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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define eprintf(...) 42
#endif

using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template<typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second

double startTime;
double getCurrentTime() {
	return ((double)clock() - startTime) / CLOCKS_PER_SEC;
}

const int N = 277;
int n;
char s[N];
vector<pii> a[N];

int main()
{
	startTime = (double)clock();
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d %s", &n, s);
//	n = 8;
//	for (int i = 0; i < (1 << n); i++)
//		s[i] = '1';
//	s[(1 << n) - 1] = '0';

	if (s[0] == '0') {
		printf("Impossible\n");
		return 0;
	}
	for (int i = 0; i < (1 << n); i++)
		for (int j = 0; j < (1 << n); j++) {
			if (s[i] == '0' && s[j] == '1' && ((i & j) == i)) {
				printf("Impossible\n");
				return 0;
			}
		}
	for (int i = 0; i < n; i++) {
		int mask = 1 << i;
		for (int j = 0; j <= i + 1; j++)
			a[mask].push_back(mp(j, 0));
		a[mask].push_back(mp(i + 1, 1));
		a[mask].push_back(mp(i, 1));
		for (int j = i; j >= 0; j--)
			a[mask].push_back(mp(j, 0));
	}
	for (int mask = 1; mask < (1 << n); mask++) {
		if (!a[mask].empty()) continue;
		int p = 0;
		while(((mask >> p) & 1) == 0) p++;
		p = 1 << p;
		int q = mask ^ p;
		a[mask].push_back(mp(0, 0));
		for (int i = 1; i < (int)a[p].size(); i++)
			a[mask].push_back(a[p][i]);
		for (int i = 1; i < (int)a[q].size(); i++)
			a[mask].push_back(a[q][i]);
		for (int i = (int)a[p].size() - 2; i >= 0; i--)
			a[mask].push_back(a[p][i]);
		for (int i = (int)a[q].size() - 2; i >= 0; i--)
			a[mask].push_back(a[q][i]);
	}
	vector<pii> ans;
	ans.push_back(mp(0, 0));
	for (int i = 0; i < (1 << n); i++) {
		if (s[i] == '1') continue;
		bool ok = true;
		for (int j = 0; j < i; j++) {
			if ((j & i) == j)
				ok &= s[j] == '1';
		}
		if (!ok) continue;
		for (int j = 1; j < (int)a[i].size(); j++)
			ans.push_back(a[i][j]);
	}
	printf("Possible\n");
	printf("%d\n", (int)ans.size() - 1);
	for (pii t : ans)
		printf("%d %d\n", t.first, t.second);

	return 0;
}

```