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 <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <chrono>
#include <climits>
#include <cmath>
#include <cstdio>
#include <deque>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <stdexcept>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;

template <class T1, class T2>
ostream& operator << (ostream& out, const pair <T1, T2> p)
{
    out << '(' << p.first << ',' << p.second << ')';
    return out;
}

template <class T1, class T2>
istream& operator >> (istream& in, pair<T1, T2> &p)
{
    in >> p.first >> p.second;
    return in;
}

template <class T>
istream& operator >> (istream& in, vector<T> &v)
{
    for (T &x : v)
        in >> x;
    return in;
}

template <class T>
ostream& operator << (ostream& out, const vector<vector<T>> &v)
{
    for (const vector<T> &x : v)
        out << x << '\n';
    return out;
}

template <class T>
ostream& operator << (ostream& out, const vector<T> &v)
{
    for (const T &x : v)
        out << x << ' ';
    return out;
}

long long gcd (long long a, long long b)
{
    if (b > a)
        swap(a, b);
    return (b ? gcd(b, a % b) : a);
}

using ll   = long long;
using pii  = pair<int, int>;
using pll  = pair<long long, long long>;
using tiii = pair<pair<int, int>, int>;
using vi   = vector<int>;
using vl   = vector<long long>;
using vvi  = vector<vector<int>>;
using vvl  = vector<vector<long long>>;

#define F          first
#define S          second
#define First      first.first
#define Second     first.second
#define Third      second
#define mp         make_pair
#define rep(i,a,b) for (int i = (a); i < (b); i++)
#define per(i,b,a) for (int i = (b); i > (a); i--)
#define all(x)     x.begin(), x.end()
#define ret(x)     return cout << x, 0;
#define throwex    throw runtime_error ("Found the error.");

const int h = 1000000007;

int n;
vector<pii> get_cycle(int mask, int cur = 0)
{
    int last_1_index = 0;
    int z = mask;
    while(z != 1)
        last_1_index++, z /= 2;
    // cerr << last_1_index << '\n';
    vector<pii> ans;
    if(cur == last_1_index)
    {
        ans.push_back({cur, 1});
        ans.push_back({cur+1, 1});
        ans.push_back({cur+1, 0});
        ans.push_back({cur, 0});
        ans.push_back({cur, 1});
        return ans;
    }
    ans.push_back({cur, 1});
    vector<pii> next_cycle = get_cycle(mask, cur + 1);
    ans.insert(ans.end(), all(next_cycle));
    ans.push_back({cur, 1});
    if(mask & (1 << cur))
    {
        ans.push_back({cur, 0});
        ans.push_back({cur+1, 0});
        reverse(all(next_cycle));
        ans.insert(ans.end(), all(next_cycle));
        ans.push_back({cur+1, 0});
        ans.push_back({cur, 0});
        ans.push_back({cur, 1});
    }
    return ans;
}

signed main()
{
    ios::sync_with_stdio(false);
    #ifdef ONLINE_JUDGE
    cin.tie(nullptr);
    cout.tie(nullptr);
    cerr.setstate(ios::failbit);
    #endif

    cin >> n;
    string a;
    cin >> a;
    rep(i,0,1 << n)
    {
        rep(j,0,i)
        {
            if((j | i) == i and a[i] == '1' and a[j] == '0')
                ret("Impossible");
        }
    }
    vector<pii> ans;
    ans.push_back({0, 1});
    rep(i,0,1 << n)
    {
        if(a[i] == '0')
        {
            vector<pii> cycle = get_cycle(i);
            ans.insert(ans.end(), cycle.begin() + 1, cycle.end());
            for(int j = i; j < (1 << n); j++)
                if((i | j) == j)
                    a[j] = '1';
            cerr << i << '\n';
            cerr << cycle << '\n';
        }
    }
    assert(ans.size() <= 250000);
    cout << "Possible\n";
    cout << ans.size() - 1 << '\n';
    for(pii x : ans)
        cout << x.F << ' ' << x.S << '\n';
}