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

int main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  string possible;
  cin >> possible;
  if (possible[0] == '0') {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 0; i < 1<<n; i++) {
    for (int j = 0; j < n; j++) {
      if (i>>j&1) {
	if (possible[i] == '1' && possible[i^1<<j] == '0') {
	  cout << "Impossible" << endl;
	  return 0;
	}
      }
    }
  }
  cout << "Possible" << endl;
  vector<pair<int,int>> ans;
  function<void(int,int)> add = [&](int mask, int inv) {
    int lsb = mask&-mask;
    if (lsb == mask) {
      int x = 31-__builtin_clz(mask);
      for (int i = 0; i <= x; i++)
	ans.emplace_back(i,0);
      if (inv) {
	ans.emplace_back(x,1);
	ans.emplace_back(x+1,1);
	ans.emplace_back(x+1,0);
      } else {
	ans.emplace_back(x+1,0);
	ans.emplace_back(x+1,1);
	ans.emplace_back(x,1);
      }
      for (int i = x; i > 0; i--)
	ans.emplace_back(i,0);
    } else {
      if (inv == 0) {
	add(mask^lsb, 0);
	add(lsb, 0);
	add(mask^lsb, 1);
	add(lsb, 1);
      } else {
	add(lsb, 0);
	add(mask^lsb, 0);
	add(lsb, 1);
	add(mask^lsb, 1);
      }
    }
  };
  for (int i = 0; i < 1<<n; i++) {
    if (possible[i] == '0') {
      add(i, 0);
    }
  }
  ans.emplace_back(0,0);
  cout << (int)ans.size()-1 << endl;
  for (auto [x,y] : ans)
    cout << x << ' ' << y << '\n';
}

### Prompt

Develop a solution in CPP 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 main() {
  ios::sync_with_stdio(0); cin.tie(0);
  int n;
  cin >> n;
  string possible;
  cin >> possible;
  if (possible[0] == '0') {
    cout << "Impossible" << endl;
    return 0;
  }
  for (int i = 0; i < 1<<n; i++) {
    for (int j = 0; j < n; j++) {
      if (i>>j&1) {
	if (possible[i] == '1' && possible[i^1<<j] == '0') {
	  cout << "Impossible" << endl;
	  return 0;
	}
      }
    }
  }
  cout << "Possible" << endl;
  vector<pair<int,int>> ans;
  function<void(int,int)> add = [&](int mask, int inv) {
    int lsb = mask&-mask;
    if (lsb == mask) {
      int x = 31-__builtin_clz(mask);
      for (int i = 0; i <= x; i++)
	ans.emplace_back(i,0);
      if (inv) {
	ans.emplace_back(x,1);
	ans.emplace_back(x+1,1);
	ans.emplace_back(x+1,0);
      } else {
	ans.emplace_back(x+1,0);
	ans.emplace_back(x+1,1);
	ans.emplace_back(x,1);
      }
      for (int i = x; i > 0; i--)
	ans.emplace_back(i,0);
    } else {
      if (inv == 0) {
	add(mask^lsb, 0);
	add(lsb, 0);
	add(mask^lsb, 1);
	add(lsb, 1);
      } else {
	add(lsb, 0);
	add(mask^lsb, 0);
	add(lsb, 1);
	add(mask^lsb, 1);
      }
    }
  };
  for (int i = 0; i < 1<<n; i++) {
    if (possible[i] == '0') {
      add(i, 0);
    }
  }
  ans.emplace_back(0,0);
  cout << (int)ans.size()-1 << endl;
  for (auto [x,y] : ans)
    cout << x << ' ' << y << '\n';
}

```