#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int N;
string s;

bool bad;

vector <int> goal (int K, string cv)
{
    //cout << K << " " << cv << "\n";
    vector <int> ans;
    if (K == 1)
    {
        if (cv[0] == '1')
        {
            if (cv[1] == '0')
            {
                ans.push_back(0);
                ans.push_back(1);
                return ans;
            }
            return ans;
        }
        if (cv[0] == '0')
        {
            if (cv[1] == '1')
                bad = true;
            // what?
            return ans;
        }
    }

    int cp = 1 << (K - 1);
    for (int i = cp; i < 2 * cp; i++)
    {
        if (cv[i] > cv[i-cp])
        {
            bad = true;
            return ans;
        }
    }

    string rv = cv.substr(0, cp);
    string lv = cv.substr(cp);
    bool l0 = true;
    for (int i = 0; i < cp; i++)
    {
        if (lv[i] == '1') l0 = false;
    }

    vector <int> rans = goal (K - 1, rv);

    if (l0)
    {
        // throw the key
        rans.push_back(K - 1);
        rans.push_back(K);
        return rans;
    }
    vector <int> lans = goal (K - 1, lv);
    swap (rans, lans);

    vector <int> god;
    for (int i : rans)
        god.push_back(i);
    god.push_back(K);
    god.push_back(K-1);
    reverse (rans.begin(), rans.end());
    for (int i : rans)
        god.push_back(i);
    god.push_back(K-1);
    god.push_back(K);
    for (int i : lans)
        god.push_back(i);
    return god;
}

int main()
{
    ios_base::sync_with_stdio(0);

    cin >> N >> s;

    bad = false;
    vector <int> res = goal (N, s);
    if (bad || s[0] == '0')
    {
        cout << "Impossible\n";
    }
    else
    {
        cout << "Possible\n";
        if (!res.size())
        {
            cout << "0\n";
            cout << "1 1\n";
            return 0;
        }
        /*for (int i : res) cout << i;
        cout << "\n";*/
        res.push_back(res[0]);

        int ctot = 0;
        for (int i = 0; i + 1 < res.size(); i++)
        {
            ctot += abs (res[i+1] - res[i]) + 1;
        }

        cout << ctot << "\n";
        for (int i = 0; i + 1 < res.size(); i++)
        {
            int cv = i % 2;
            int nv = 1 - cv;
            cout << res[i] << " " << cv << "\n";
            for (int j = res[i]; j < res[i+1]; j++)
                cout << j << " " << nv << "\n";
            for (int j = res[i]; j > res[i+1]; j--)
                cout << j << " " << nv << "\n";
        }
        cout << res[0] << " " << 0 << "\n";
    }
}
### Prompt

Please provide a Cpp coded solution 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;
typedef long long ll;

int N;
string s;

bool bad;

vector <int> goal (int K, string cv)
{
    //cout << K << " " << cv << "\n";
    vector <int> ans;
    if (K == 1)
    {
        if (cv[0] == '1')
        {
            if (cv[1] == '0')
            {
                ans.push_back(0);
                ans.push_back(1);
                return ans;
            }
            return ans;
        }
        if (cv[0] == '0')
        {
            if (cv[1] == '1')
                bad = true;
            // what?
            return ans;
        }
    }

    int cp = 1 << (K - 1);
    for (int i = cp; i < 2 * cp; i++)
    {
        if (cv[i] > cv[i-cp])
        {
            bad = true;
            return ans;
        }
    }

    string rv = cv.substr(0, cp);
    string lv = cv.substr(cp);
    bool l0 = true;
    for (int i = 0; i < cp; i++)
    {
        if (lv[i] == '1') l0 = false;
    }

    vector <int> rans = goal (K - 1, rv);

    if (l0)
    {
        // throw the key
        rans.push_back(K - 1);
        rans.push_back(K);
        return rans;
    }
    vector <int> lans = goal (K - 1, lv);
    swap (rans, lans);

    vector <int> god;
    for (int i : rans)
        god.push_back(i);
    god.push_back(K);
    god.push_back(K-1);
    reverse (rans.begin(), rans.end());
    for (int i : rans)
        god.push_back(i);
    god.push_back(K-1);
    god.push_back(K);
    for (int i : lans)
        god.push_back(i);
    return god;
}

int main()
{
    ios_base::sync_with_stdio(0);

    cin >> N >> s;

    bad = false;
    vector <int> res = goal (N, s);
    if (bad || s[0] == '0')
    {
        cout << "Impossible\n";
    }
    else
    {
        cout << "Possible\n";
        if (!res.size())
        {
            cout << "0\n";
            cout << "1 1\n";
            return 0;
        }
        /*for (int i : res) cout << i;
        cout << "\n";*/
        res.push_back(res[0]);

        int ctot = 0;
        for (int i = 0; i + 1 < res.size(); i++)
        {
            ctot += abs (res[i+1] - res[i]) + 1;
        }

        cout << ctot << "\n";
        for (int i = 0; i + 1 < res.size(); i++)
        {
            int cv = i % 2;
            int nv = 1 - cv;
            cout << res[i] << " " << cv << "\n";
            for (int j = res[i]; j < res[i+1]; j++)
                cout << j << " " << nv << "\n";
            for (int j = res[i]; j > res[i+1]; j--)
                cout << j << " " << nv << "\n";
        }
        cout << res[0] << " " << 0 << "\n";
    }
}
```