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

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n;
    string s;
    cin >> n >> s;
    if (s[0] == '0') {
        cout << "Impossible\n";
        return 0;
    }
    vector<bool> z(1 << n);
    vector<pair<int, int>> p;
    p.push_back({0, 0});
    for (int j = 1; j < (1 << n); j++) {
        if (z[j]) {
            if (s[j] == '1') {
                cout << "Impossible\n";
                return 0;
            }
            for (int i = 0; i < n; i++) {
                z[j | (1 << i)] = true;
            }
            continue;
        }
        if (s[j] == '0') {
            vector<pair<int, int>> pp;
            for (int i = 0; i < n; i++) {
                if (j & (1 << i)) {
                    if (pp.size() == 0) {
                        for (int x = 0; x <= i; x++) {
                            pp.push_back({x, 0});
                        }
                        pp.push_back({i, 1});
                        pp.push_back({i + 1, 1});
                        for (int x = i + 1; x >= 0; x--) {
                            pp.push_back({x, 0});
                        }
                    } else {
                        vector<pair<int, int>> ppp = pp;
                        for (int x = 1; x <= i; x++) {
                            ppp.push_back({x, 0});
                        }
                        ppp.push_back({i, 1});
                        ppp.push_back({i + 1, 1});
                        for (int x = i + 1; x >= 1; x--) {
                            ppp.push_back({x, 0});
                        }
                        reverse(pp.begin(), pp.end());
                        ppp.insert(ppp.end(), pp.begin(), pp.end());
                        for (int x = 1; x <= i + 1; x++) {
                            ppp.push_back({x, 0});
                        }
                        ppp.push_back({i + 1, 1});
                        ppp.push_back({i, 1});
                        for (int x = i; x >= 0; x--) {
                            ppp.push_back({x, 0});
                        }
                        pp = ppp;
                    }
                }
            }
            p.insert(p.end(), pp.begin() + 1, pp.end());
            for (int i = 0; i < n; i++) {
                z[j | (1 << i)] = true;
            }
        }
    }
    bool ok = false;
    while (!ok) {
        vector<pair<int, int>> pp;
        ok = true;
        for (int i = 0; i < p.size(); i++) {
            if (i > 0 && i < (int)p.size() - 1 && p[i - 1] == p[i + 1]) {
                ok = false;
                i++;
            } else {
                pp.push_back(p[i]);
            }
        }
        p = pp;
    }
    cout << "Possible\n";
    cout << p.size() - 1 << "\n";
    for (auto x : p) {
        cout << x.first << " " << x.second << "\n";
    }

    return 0;
}
### Prompt

Please create a solution in CPP to 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 <bits/stdc++.h>
#define long long long int
using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n;
    string s;
    cin >> n >> s;
    if (s[0] == '0') {
        cout << "Impossible\n";
        return 0;
    }
    vector<bool> z(1 << n);
    vector<pair<int, int>> p;
    p.push_back({0, 0});
    for (int j = 1; j < (1 << n); j++) {
        if (z[j]) {
            if (s[j] == '1') {
                cout << "Impossible\n";
                return 0;
            }
            for (int i = 0; i < n; i++) {
                z[j | (1 << i)] = true;
            }
            continue;
        }
        if (s[j] == '0') {
            vector<pair<int, int>> pp;
            for (int i = 0; i < n; i++) {
                if (j & (1 << i)) {
                    if (pp.size() == 0) {
                        for (int x = 0; x <= i; x++) {
                            pp.push_back({x, 0});
                        }
                        pp.push_back({i, 1});
                        pp.push_back({i + 1, 1});
                        for (int x = i + 1; x >= 0; x--) {
                            pp.push_back({x, 0});
                        }
                    } else {
                        vector<pair<int, int>> ppp = pp;
                        for (int x = 1; x <= i; x++) {
                            ppp.push_back({x, 0});
                        }
                        ppp.push_back({i, 1});
                        ppp.push_back({i + 1, 1});
                        for (int x = i + 1; x >= 1; x--) {
                            ppp.push_back({x, 0});
                        }
                        reverse(pp.begin(), pp.end());
                        ppp.insert(ppp.end(), pp.begin(), pp.end());
                        for (int x = 1; x <= i + 1; x++) {
                            ppp.push_back({x, 0});
                        }
                        ppp.push_back({i + 1, 1});
                        ppp.push_back({i, 1});
                        for (int x = i; x >= 0; x--) {
                            ppp.push_back({x, 0});
                        }
                        pp = ppp;
                    }
                }
            }
            p.insert(p.end(), pp.begin() + 1, pp.end());
            for (int i = 0; i < n; i++) {
                z[j | (1 << i)] = true;
            }
        }
    }
    bool ok = false;
    while (!ok) {
        vector<pair<int, int>> pp;
        ok = true;
        for (int i = 0; i < p.size(); i++) {
            if (i > 0 && i < (int)p.size() - 1 && p[i - 1] == p[i + 1]) {
                ok = false;
                i++;
            } else {
                pp.push_back(p[i]);
            }
        }
        p = pp;
    }
    cout << "Possible\n";
    cout << p.size() - 1 << "\n";
    for (auto x : p) {
        cout << x.first << " " << x.second << "\n";
    }

    return 0;
}
```