#include <bits/stdc++.h>
using namespace std;

template <typename T>
vector<T>& operator += (vector<T>& a, vector<T> b) {
	for (auto x : b) {
		a.push_back(x);
	}
	return a;
}
template <typename T>
vector<T> operator ~ (vector<T> a) {
  reverse(a.begin(), a.end());
  return a;
}

typedef class Point {
  public:
    int x, y;

    Point() { }
    Point(int x, int y) : x(x), y(y) {  }

    bool operator == (Point b) {
      return x == b.x && y == b.y;
    }
} Point;

typedef class Data {
  public:
    char dir;
    int num;

    Data() {  }
    Data(char dir, int num) : dir(dir), num(num) {  }
} Data;

int n, N;

vector<Point> _export(vector<Data> vd) {
  vector<Point> ret {Point(n, 1)};
  for (auto t : vd) {
    auto lst = ret.back();
    if (t.dir == 'd' && lst.y == 1) {
      ret.emplace_back(lst.x, 0);
    } else if (t.dir == 'u' && lst.y == 0) {
      ret.emplace_back(lst.x, 1);
    }
    lst = ret.back();
    if (lst.x == t.num) {
      lst.x--;
    } else if (lst.x == t.num - 1) {
      lst.x++;
    } else {
      assert(false);
    }
    ret.push_back(lst);
  }
  if (ret[0] == ret.back()) {
    ret.pop_back();
  }
  return ret;
}
vector<Point> _export(vector<Data> vd, vector<int> pos) {
  for (auto& t : vd) {
    t.num = pos[t.num - 1] + 1;
  }
  vector<Data> nvd;
  for (int i = n; i > vd[0].num; i--) {
      nvd.emplace_back('d', i);
  }
  for (auto t : vd) {
    if (!nvd.empty()) {
      auto& lst = nvd.back();
      for (int i = lst.num + 1; i < t.num; i++) {
        nvd.emplace_back('d', i);
      }
      for (int i = lst.num - 1; i > t.num; i--) {
        nvd.emplace_back('d', i);
      }
    }
    nvd.push_back(t);
  }
  for (int i = nvd.back().num + 1; i <= n; i++) {
    nvd.emplace_back('d', i);
  }
  return _export(nvd);
}

char A[530];
vector<Data> df[10];

#define NO do puts("Impossible"), exit(0); while (0)

int main() {
  scanf("%d", &n);
  N = 1 << n;
  scanf("%s", A);
  if (A[0] != '1') {
    NO;
  }
	for (int s = 0; s < N; s++) {
    if (A[s] == '1') {
      for (int i = 0; i < n; i++) {
        if (((s >> i) & 1) && A[s ^ (1 << i)] == '0') {
          NO;
        }
      }
    }
  }
  df[1].emplace_back('d', 1);
  df[1].emplace_back('u', 1);
  for (int i = 2; i <= n; i++) {
    df[i].emplace_back('u', i);
    df[i] += df[i - 1];
    df[i].emplace_back('u', i);
    df[i].emplace_back('d', i);
    df[i] += ~df[i - 1];
    df[i].emplace_back('d', i);
  }
  vector<Point> ans;
  for (int s = 1; s < N; s++) {
    if (A[s] == '0') {
      vector<int> pos;
      int bit = 0;
      for (int i = 0; i < n; i++) {
        if ((s >> i) & 1) {
          pos.push_back(i);
          bit++;
        }
      }
      ans += _export(df[bit], pos);
    }
  }
  puts("Possible");
  printf("%d\n", (signed) ans.size());
  if (ans.empty()) {
    ans.emplace_back(0, 0);
  } else {
    ans.push_back(ans[0]);
  }
  for (auto p : ans) {
    printf("%d %d\n", p.x, p.y);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;

template <typename T>
vector<T>& operator += (vector<T>& a, vector<T> b) {
	for (auto x : b) {
		a.push_back(x);
	}
	return a;
}
template <typename T>
vector<T> operator ~ (vector<T> a) {
  reverse(a.begin(), a.end());
  return a;
}

typedef class Point {
  public:
    int x, y;

    Point() { }
    Point(int x, int y) : x(x), y(y) {  }

    bool operator == (Point b) {
      return x == b.x && y == b.y;
    }
} Point;

typedef class Data {
  public:
    char dir;
    int num;

    Data() {  }
    Data(char dir, int num) : dir(dir), num(num) {  }
} Data;

int n, N;

vector<Point> _export(vector<Data> vd) {
  vector<Point> ret {Point(n, 1)};
  for (auto t : vd) {
    auto lst = ret.back();
    if (t.dir == 'd' && lst.y == 1) {
      ret.emplace_back(lst.x, 0);
    } else if (t.dir == 'u' && lst.y == 0) {
      ret.emplace_back(lst.x, 1);
    }
    lst = ret.back();
    if (lst.x == t.num) {
      lst.x--;
    } else if (lst.x == t.num - 1) {
      lst.x++;
    } else {
      assert(false);
    }
    ret.push_back(lst);
  }
  if (ret[0] == ret.back()) {
    ret.pop_back();
  }
  return ret;
}
vector<Point> _export(vector<Data> vd, vector<int> pos) {
  for (auto& t : vd) {
    t.num = pos[t.num - 1] + 1;
  }
  vector<Data> nvd;
  for (int i = n; i > vd[0].num; i--) {
      nvd.emplace_back('d', i);
  }
  for (auto t : vd) {
    if (!nvd.empty()) {
      auto& lst = nvd.back();
      for (int i = lst.num + 1; i < t.num; i++) {
        nvd.emplace_back('d', i);
      }
      for (int i = lst.num - 1; i > t.num; i--) {
        nvd.emplace_back('d', i);
      }
    }
    nvd.push_back(t);
  }
  for (int i = nvd.back().num + 1; i <= n; i++) {
    nvd.emplace_back('d', i);
  }
  return _export(nvd);
}

char A[530];
vector<Data> df[10];

#define NO do puts("Impossible"), exit(0); while (0)

int main() {
  scanf("%d", &n);
  N = 1 << n;
  scanf("%s", A);
  if (A[0] != '1') {
    NO;
  }
	for (int s = 0; s < N; s++) {
    if (A[s] == '1') {
      for (int i = 0; i < n; i++) {
        if (((s >> i) & 1) && A[s ^ (1 << i)] == '0') {
          NO;
        }
      }
    }
  }
  df[1].emplace_back('d', 1);
  df[1].emplace_back('u', 1);
  for (int i = 2; i <= n; i++) {
    df[i].emplace_back('u', i);
    df[i] += df[i - 1];
    df[i].emplace_back('u', i);
    df[i].emplace_back('d', i);
    df[i] += ~df[i - 1];
    df[i].emplace_back('d', i);
  }
  vector<Point> ans;
  for (int s = 1; s < N; s++) {
    if (A[s] == '0') {
      vector<int> pos;
      int bit = 0;
      for (int i = 0; i < n; i++) {
        if ((s >> i) & 1) {
          pos.push_back(i);
          bit++;
        }
      }
      ans += _export(df[bit], pos);
    }
  }
  puts("Possible");
  printf("%d\n", (signed) ans.size());
  if (ans.empty()) {
    ans.emplace_back(0, 0);
  } else {
    ans.push_back(ans[0]);
  }
  for (auto p : ans) {
    printf("%d %d\n", p.x, p.y);
  }
  return 0;
}

```