#include <iostream>
#include <vector>
#include <cassert>

using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
template <class T> using V = vector<T>;
template <class T> using VV = V<V<T>>;

int main() {
    int n;
    cin >> n;
    string _s;
    cin >> _s;
    V<bool> s;
    for (char c: _s) s.push_back(c == '1');
    for (int i = 0; i < (1 << n); i++) {
        for (int j = 0; j < n; j++) {
            if (!s[i] && s[i | (1 << j)]) {                
                cout << "Impossible" << endl;
                return 0;
            }
        }
    }

    using P = pair<int, int>;
    V<P> ans;
    for (int i = 0; i < (1 << n); i++) {
        if (s[i]) continue;
        bool minimum = true;
        for (int j = 0; j < n; j++) {
            if ((i & (1 << j)) && !s[i ^ (1 << j)]) {
                minimum = false;
                break;
            }
        }
        if (!minimum) continue;
        //
        int st = 0;
        while (!(i & (1 << st))) st++;
        V<P> path = {P(st + 1, 0), P(st, 0), P(st, 1), P(st + 1, 1)};
        for (int j = st + 1; j < n; j++) {
            if (i & (1 << j)) {
                V<P> npath = {P(j + 1, 0)};
                npath.insert(npath.end(), path.begin(), path.end());
                npath.push_back(P(j, 0));
                npath.push_back(P(j + 1, 0));
                npath.push_back(P(j + 1, 1));
                npath.insert(npath.end(), path.rbegin(), path.rend());
                npath.push_back(P(j, 1));
                npath.push_back(P(j + 1, 1));
                path = npath;
            } else {
                path.insert(path.begin(), P(j + 1, 0));
                path.push_back(P(j, 0));
                path.push_back(P(j + 1, 0));
                path.push_back(P(j + 1, 1));
            }
        }
        ans.insert(ans.end(), path.begin(), path.end());
    }
    ans.push_back(P(n, 0));
/*    V<P> ans2;
    for (auto p: ans) {
        ans2.push_back(p);
        int s = int(ans2.size());
        if (s >= 3 && ans2[s - 1] == ans2[s - 3]) {
            ans2.pop_back();
            ans2.pop_back();
        }
    }
    ans = ans2;*/

    assert(ans.size() <= 250000);
    cout << "Possible" << "\n";
    cout << int(ans.size()) - 1 << "\n";
    for (auto p: ans) {
        cout << p.first << " " << p.second << "\n";
    }
    return 0;
}

### Prompt

Create a solution in CPP for the following problem:

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 <iostream>
#include <vector>
#include <cassert>

using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n - 1); }
template <class T> using V = vector<T>;
template <class T> using VV = V<V<T>>;

int main() {
    int n;
    cin >> n;
    string _s;
    cin >> _s;
    V<bool> s;
    for (char c: _s) s.push_back(c == '1');
    for (int i = 0; i < (1 << n); i++) {
        for (int j = 0; j < n; j++) {
            if (!s[i] && s[i | (1 << j)]) {                
                cout << "Impossible" << endl;
                return 0;
            }
        }
    }

    using P = pair<int, int>;
    V<P> ans;
    for (int i = 0; i < (1 << n); i++) {
        if (s[i]) continue;
        bool minimum = true;
        for (int j = 0; j < n; j++) {
            if ((i & (1 << j)) && !s[i ^ (1 << j)]) {
                minimum = false;
                break;
            }
        }
        if (!minimum) continue;
        //
        int st = 0;
        while (!(i & (1 << st))) st++;
        V<P> path = {P(st + 1, 0), P(st, 0), P(st, 1), P(st + 1, 1)};
        for (int j = st + 1; j < n; j++) {
            if (i & (1 << j)) {
                V<P> npath = {P(j + 1, 0)};
                npath.insert(npath.end(), path.begin(), path.end());
                npath.push_back(P(j, 0));
                npath.push_back(P(j + 1, 0));
                npath.push_back(P(j + 1, 1));
                npath.insert(npath.end(), path.rbegin(), path.rend());
                npath.push_back(P(j, 1));
                npath.push_back(P(j + 1, 1));
                path = npath;
            } else {
                path.insert(path.begin(), P(j + 1, 0));
                path.push_back(P(j, 0));
                path.push_back(P(j + 1, 0));
                path.push_back(P(j + 1, 1));
            }
        }
        ans.insert(ans.end(), path.begin(), path.end());
    }
    ans.push_back(P(n, 0));
/*    V<P> ans2;
    for (auto p: ans) {
        ans2.push_back(p);
        int s = int(ans2.size());
        if (s >= 3 && ans2[s - 1] == ans2[s - 3]) {
            ans2.pop_back();
            ans2.pop_back();
        }
    }
    ans = ans2;*/

    assert(ans.size() <= 250000);
    cout << "Possible" << "\n";
    cout << int(ans.size()) - 1 << "\n";
    for (auto p: ans) {
        cout << p.first << " " << p.second << "\n";
    }
    return 0;
}

```