#include <bits/stdc++.h>

using namespace std;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

int n, x, y;
bool a[1 << 9];
vector<pair<int, int>> res;

void move(int tx, int ty) {
  while (x > tx)
    x--, res.push_back({x, y});
  while (x < tx)
    x++, res.push_back({x, y});
  while (y > ty)
    y--, res.push_back({x, y});
  while (y < ty)
    y++, res.push_back({x, y});
}

void solve(vector<int> p, bool d = false) {
  if (sz(p) == 1) {
    move(p[0] + d, 0);
    move(p[0] + d, 1);
    move(p[0] + !d, 1);
    move(p[0] + !d, 0);
    return;
  }
  vector<int> c;
  c.push_back(p.back());
  p.pop_back();
  if (!d) {
    solve(c, d);
    solve(p, d);
    solve(c, !d);
    solve(p, !d);
  } else {
    solve(p, !d);
    solve(c, !d);
    solve(p, d);
    solve(c, d);
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  char c;
  bool p = true;
  for (int i = 0; i < 1 << n; i++) {
    cin >> c, a[i] = c == '1';
    if (a[i])
      for (int j = 1 << n; j >= 0; j--)
        p &= a[i & j];
  }
  p &= a[0];
  if (!p) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 0; i < 1 << n; i++) {
    p = !a[i];
    for (int j = 1 << n; j >= 0; j--)
      if ((i & j) != i)
        p &= a[i & j];
    if (p) {
      vector<int> pos;
      for (int j = n - 1; j >= 0; j--)
        if ((i >> j) & 1)
          pos.push_back(j);
      solve(pos);
    }
  }
  move(0, 0);
  cout << "Possible\n";
  cout << sz(res) << "\n";
  cout << "0 0\n";
  for (auto pos : res)
    cout << pos.first << " " << pos.second << "\n";
}

### 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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

int n, x, y;
bool a[1 << 9];
vector<pair<int, int>> res;

void move(int tx, int ty) {
  while (x > tx)
    x--, res.push_back({x, y});
  while (x < tx)
    x++, res.push_back({x, y});
  while (y > ty)
    y--, res.push_back({x, y});
  while (y < ty)
    y++, res.push_back({x, y});
}

void solve(vector<int> p, bool d = false) {
  if (sz(p) == 1) {
    move(p[0] + d, 0);
    move(p[0] + d, 1);
    move(p[0] + !d, 1);
    move(p[0] + !d, 0);
    return;
  }
  vector<int> c;
  c.push_back(p.back());
  p.pop_back();
  if (!d) {
    solve(c, d);
    solve(p, d);
    solve(c, !d);
    solve(p, !d);
  } else {
    solve(p, !d);
    solve(c, !d);
    solve(p, d);
    solve(c, d);
  }
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n;
  char c;
  bool p = true;
  for (int i = 0; i < 1 << n; i++) {
    cin >> c, a[i] = c == '1';
    if (a[i])
      for (int j = 1 << n; j >= 0; j--)
        p &= a[i & j];
  }
  p &= a[0];
  if (!p) {
    cout << "Impossible\n";
    return 0;
  }
  for (int i = 0; i < 1 << n; i++) {
    p = !a[i];
    for (int j = 1 << n; j >= 0; j--)
      if ((i & j) != i)
        p &= a[i & j];
    if (p) {
      vector<int> pos;
      for (int j = n - 1; j >= 0; j--)
        if ((i >> j) & 1)
          pos.push_back(j);
      solve(pos);
    }
  }
  move(0, 0);
  cout << "Possible\n";
  cout << sz(res) << "\n";
  cout << "0 0\n";
  for (auto pos : res)
    cout << pos.first << " " << pos.second << "\n";
}

```