/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include <iostream>
#include <fstream>

#include <bits/stdc++.h>

#define pb push_back
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(),(v).end()
#define mp make_pair

using namespace std;

typedef long long int64;
typedef vector<int> vi;
typedef pair<int, int> ii;

class TaskE {
 public:
  string inverse(string s) {
    string res = "";
    for (int i = s.size() - 1; i >= 0; --i) {
      char ch = s[i];
      if (ch >= 'a' && ch <= 'z') {
        ch += 'A' - 'a';
      } else {
        ch += 'a' - 'A';
      }
      res += ch;
    }
    return res;
  }

  void solveOne(istream &in, ostream &out) {
    int n;
    in >> n;
    string a;
    in >> a;
    vector<bool> tangled(1 << n);
    for (int i = 0; i < (1 << n); ++i) {
      tangled[i] = a[i] == '0';
    }
    if (tangled[0]) {
      out << "Impossible\n";
      return;
    }
    for (int s1 = 0; s1 < (1 << n); ++s1) {
      for (int s2 = 0; s2 < (1 << n); ++s2) {
        if ((s1 & s2) == s1 && !tangled[s2] && tangled[s1]) {
          out << "Impossible\n";
          return;
        }
      }
    }
    vector<ii> pts;
    pts.emplace_back(0, 0);
    for (int s1 = 0; s1 < (1 << n); ++s1) {
      if (tangled[s1]) {
        bool ok = true;
        for (int s2 = 0; s2 < (1 << n); ++s2) {
          if (s1 != s2 && ((s1 & s2) == s2) && tangled[s2]) {
            ok = false;
          }
        }
        if (!ok) continue;
        string seq = "";
        for (int i = 0; i < n; ++i) {
          if ((s1 & (1 << i)) != 0) {
            char ch = 'a' + i;
            if (seq.empty()) {
              seq += ch;
            } else {
              seq = ch + seq + (char) (ch + 'A' - 'a') + inverse(seq);
            }
          }
        }
        cerr << seq << endl;
        int x = 0;
        for (int p = 0; p < seq.size(); ++p) {
          char ch = seq[p];
          int i;
          int dir;
          if (ch >= 'A' && ch <= 'Z') {
            i = ch - 'A';
            dir = 0;
          } else {
            i = ch - 'a';
            dir = 1;
          }
          if (x < i) {
            for (int j = x + 1; j <= i; ++j) {
              pts.emplace_back(j, 0);
            }
          } else if (x > i) {
            for (int j = x - 1; j >= i; --j) {
              pts.emplace_back(j, 0);
            }
          }
          x = i;
          if (dir == 0) {
            pts.emplace_back(i + 1, 0);
            pts.emplace_back(i + 1, 1);
            pts.emplace_back(i, 1);
            pts.emplace_back(i, 0);
          } else {
            pts.emplace_back(i, 1);
            pts.emplace_back(i + 1, 1);
            pts.emplace_back(i + 1, 0);
            pts.emplace_back(i, 0);
          }
        }
        for (int j = x - 1; j >= 0; --j) {
          pts.emplace_back(j, 0);
        }
        x = 0;
      }
    }
    out << "Possible\n";
    out << pts.size() - 1 << "\n";
    for (const auto& p : pts) {
      out << p.first << " " << p.second << "\n";
    }
  }

  void solve(std::istream &in, std::ostream &out) {
    int nt;
    nt = 1;
    for (int it = 0; it < nt; ++it) {
      solveOne(in, out);
    }
  }
};


int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  TaskE solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

### 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
/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author
 */

#include <iostream>
#include <fstream>

#include <bits/stdc++.h>

#define pb push_back
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(),(v).end()
#define mp make_pair

using namespace std;

typedef long long int64;
typedef vector<int> vi;
typedef pair<int, int> ii;

class TaskE {
 public:
  string inverse(string s) {
    string res = "";
    for (int i = s.size() - 1; i >= 0; --i) {
      char ch = s[i];
      if (ch >= 'a' && ch <= 'z') {
        ch += 'A' - 'a';
      } else {
        ch += 'a' - 'A';
      }
      res += ch;
    }
    return res;
  }

  void solveOne(istream &in, ostream &out) {
    int n;
    in >> n;
    string a;
    in >> a;
    vector<bool> tangled(1 << n);
    for (int i = 0; i < (1 << n); ++i) {
      tangled[i] = a[i] == '0';
    }
    if (tangled[0]) {
      out << "Impossible\n";
      return;
    }
    for (int s1 = 0; s1 < (1 << n); ++s1) {
      for (int s2 = 0; s2 < (1 << n); ++s2) {
        if ((s1 & s2) == s1 && !tangled[s2] && tangled[s1]) {
          out << "Impossible\n";
          return;
        }
      }
    }
    vector<ii> pts;
    pts.emplace_back(0, 0);
    for (int s1 = 0; s1 < (1 << n); ++s1) {
      if (tangled[s1]) {
        bool ok = true;
        for (int s2 = 0; s2 < (1 << n); ++s2) {
          if (s1 != s2 && ((s1 & s2) == s2) && tangled[s2]) {
            ok = false;
          }
        }
        if (!ok) continue;
        string seq = "";
        for (int i = 0; i < n; ++i) {
          if ((s1 & (1 << i)) != 0) {
            char ch = 'a' + i;
            if (seq.empty()) {
              seq += ch;
            } else {
              seq = ch + seq + (char) (ch + 'A' - 'a') + inverse(seq);
            }
          }
        }
        cerr << seq << endl;
        int x = 0;
        for (int p = 0; p < seq.size(); ++p) {
          char ch = seq[p];
          int i;
          int dir;
          if (ch >= 'A' && ch <= 'Z') {
            i = ch - 'A';
            dir = 0;
          } else {
            i = ch - 'a';
            dir = 1;
          }
          if (x < i) {
            for (int j = x + 1; j <= i; ++j) {
              pts.emplace_back(j, 0);
            }
          } else if (x > i) {
            for (int j = x - 1; j >= i; --j) {
              pts.emplace_back(j, 0);
            }
          }
          x = i;
          if (dir == 0) {
            pts.emplace_back(i + 1, 0);
            pts.emplace_back(i + 1, 1);
            pts.emplace_back(i, 1);
            pts.emplace_back(i, 0);
          } else {
            pts.emplace_back(i, 1);
            pts.emplace_back(i + 1, 1);
            pts.emplace_back(i + 1, 0);
            pts.emplace_back(i, 0);
          }
        }
        for (int j = x - 1; j >= 0; --j) {
          pts.emplace_back(j, 0);
        }
        x = 0;
      }
    }
    out << "Possible\n";
    out << pts.size() - 1 << "\n";
    for (const auto& p : pts) {
      out << p.first << " " << p.second << "\n";
    }
  }

  void solve(std::istream &in, std::ostream &out) {
    int nt;
    nt = 1;
    for (int it = 0; it < nt; ++it) {
      solveOne(in, out);
    }
  }
};


int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  TaskE solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}

```