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;
const int MAXN = 15;
const int MAXS = 1 << 8;
typedef long long ll;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
vector <pair <int, int>> ans, tmp;
int n, bit[MAXN]; bool flg[MAXS]; char s[MAXS];
void augment(vector <pair <int, int>> &res, int x) {
	if (res.size() == 0) {
		res.emplace_back(x, 0);
		res.emplace_back(x, 1);
		return;
	}
	vector <pair <int, int>> tmp;
	tmp.emplace_back(x, 0);
	for (auto x : res) tmp.push_back(x);
	tmp.emplace_back(x, 0);
	reverse(tmp.begin(), tmp.end());
	tmp.emplace_back(x, 1);
	for (auto x : res) tmp.push_back(x);
	tmp.emplace_back(x, 1);
	res = tmp;
}
void work(int mask) {
	tmp.clear(); int cur = 0;
	for (int i = n; i >= 1; i--)
		if (mask & bit[i]) augment(tmp, i);
	for (auto x : tmp) {
		if (x.second == 0) {
			if (cur >= x.first) while (cur >= x.first) ans.emplace_back(--cur, 0);
			else while (cur < x.first) ans.emplace_back(++cur, 0);
		} else {
			if (cur >= x.first) {
				while (cur >= x.first + 1) ans.emplace_back(--cur, 0);
				ans.emplace_back(cur, 1);
				ans.emplace_back(--cur, 1);
				ans.emplace_back(cur, 0);
			} else {
				while (cur < x.first - 1) ans.emplace_back(++cur, 0);
				ans.emplace_back(cur, 1);
				ans.emplace_back(++cur, 1);
				ans.emplace_back(cur, 0);
			}
		}
	}
	while (cur > 0) ans.emplace_back(--cur, 0);
}
int main() {
	read(n), scanf("\n%s", s);
	for (int i = 1; i <= n; i++)
		bit[i] = 1 << (i - 1);
	int goal = 1 << n;
	for (int i = 0; i < goal; i++)
	for (int j = 0; j < i; j++)
		if ((i & j) == j && s[j] == '0') flg[i] = true;
	for (int i = 0; i < goal; i++)
		if (s[i] == '1' && flg[i]) {
			puts("Impossible");
			return 0;
		}
	puts("Possible");
	for (int i = 0; i < goal; i++)
		if (s[i] == '0' && !flg[i]) work(i);
	cout << ans.size() << endl;
	printf("%d %d\n", 0, 0);
	for (auto x : ans)
		printf("%d %d\n", x.first, x.second);
	return 0;
}