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

void f(stringstream &ss, int r, int &j, int &l) {
    int s = abs(r);
    while(s < j)
        ss << --j << " 0\n", l++;
    while(s > j)
        ss << ++j << " 0\n", l++;
    if(r < 0)
        ss << s << " 1\n"
           << s - 1 << " 1\n"
           << s - 1 << " 0\n",
            j = s - 1, l += 3;
    else
        ss << s << " 1\n"
           << s + 1 << " 1\n"
           << s + 1 << " 0\n",
            j = s + 1, l += 3;
}
vector<int> g(int k) {
    int i = 0;
    while(!(k & 1 << i))
        i++;
    vector<int> v;
    v.push_back(i);
    if(k == 1 << i)
        return v;
    auto w = g(k ^ 1 << i);
    v.insert(v.end(), w.begin(), w.end());
    v.push_back(-1 - i);
    for(int &l : w)
        l = -1 - l;
    reverse(w.begin(), w.end());
    v.insert(v.end(), w.begin(), w.end());
    return v;
}
int main() {
    int n;
    string a;
    cin >> n >> a;
    vector<int> b;
    for(int i = 0; i < 1 << n; i++) {
        int x = 0;
        for(int &k : b) {
            if((k & i) == k) {
                if(a[i] == '1') {
                    cout << "Impossible" << endl;
                    return 0;
                }
                x = 1;
            }
        }
        if(a[i] == '0' && !x)
            b.push_back(i);
    }
    cout << "Possible\n";
    stringstream ss;
    int l = 0, j = 0;
    ss << "0 0\n";
    for(int k : b) {
        auto v = g(k);
        for(int i : v)
            f(ss, i, j, l);
    }
    if(j)
        for(j--; j >= 0; j--)
            ss << j << " 0\n", l++;
    cout << l << '\n' << ss.str() << flush;
    return 0;
}

### Prompt

Your challenge is to write a cpp solution 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>
using namespace std;

void f(stringstream &ss, int r, int &j, int &l) {
    int s = abs(r);
    while(s < j)
        ss << --j << " 0\n", l++;
    while(s > j)
        ss << ++j << " 0\n", l++;
    if(r < 0)
        ss << s << " 1\n"
           << s - 1 << " 1\n"
           << s - 1 << " 0\n",
            j = s - 1, l += 3;
    else
        ss << s << " 1\n"
           << s + 1 << " 1\n"
           << s + 1 << " 0\n",
            j = s + 1, l += 3;
}
vector<int> g(int k) {
    int i = 0;
    while(!(k & 1 << i))
        i++;
    vector<int> v;
    v.push_back(i);
    if(k == 1 << i)
        return v;
    auto w = g(k ^ 1 << i);
    v.insert(v.end(), w.begin(), w.end());
    v.push_back(-1 - i);
    for(int &l : w)
        l = -1 - l;
    reverse(w.begin(), w.end());
    v.insert(v.end(), w.begin(), w.end());
    return v;
}
int main() {
    int n;
    string a;
    cin >> n >> a;
    vector<int> b;
    for(int i = 0; i < 1 << n; i++) {
        int x = 0;
        for(int &k : b) {
            if((k & i) == k) {
                if(a[i] == '1') {
                    cout << "Impossible" << endl;
                    return 0;
                }
                x = 1;
            }
        }
        if(a[i] == '0' && !x)
            b.push_back(i);
    }
    cout << "Possible\n";
    stringstream ss;
    int l = 0, j = 0;
    ss << "0 0\n";
    for(int k : b) {
        auto v = g(k);
        for(int i : v)
            f(ss, i, j, l);
    }
    if(j)
        for(j--; j >= 0; j--)
            ss << j << " 0\n", l++;
    cout << l << '\n' << ss.str() << flush;
    return 0;
}

```