#include <cstdio>
#include <vector>
#include <utility>

using namespace std;

vector<pair<int, int> > ans;

void cw(int loc)
{
    for (int i = 1; i <= loc - 1; i++) {
        ans.push_back(make_pair(i, 0));
    }
    ans.push_back(make_pair(loc-1, 1));
    ans.push_back(make_pair(loc, 1));
    for (int i = loc; i >= 0; i--) {
        ans.push_back(make_pair(i, 0));
    }
}

void ccw(int loc)
{
    for (int i = 1; i <= loc; i++) {
        ans.push_back(make_pair(i, 0));
    }
    ans.push_back(make_pair(loc, 1));
    ans.push_back(make_pair(loc-1, 1));
    for (int i = loc - 1; i >= 0; i--) {
        ans.push_back(make_pair(i, 0));
    }
}

void work(vector<int>& elems, vector<int>& ops)
{
    if (elems.size() == 1) {
        ops.push_back(elems[0]); return;
    }
    vector<int> left;
    vector<int> right;
    for (int i = 0; i < elems.size() / 2; i++) {
        left.push_back(elems[i]);
    }
    for (int i = elems.size() / 2; i < elems.size(); i++) {
        right.push_back(elems[i]);
    }
    vector<int> leftans;
    vector<int> rightans;
    work(left, leftans);
    work(right, rightans);
    for (int i = 0; i < leftans.size(); i++) {
        ops.push_back(leftans[i]);
    }
    for (int i = 0; i < rightans.size(); i++) {
        ops.push_back(rightans[i]);
    }
    reverse(leftans.begin(), leftans.end());
    reverse(rightans.begin(), rightans.end());
    for (int i = 0; i < leftans.size(); i++) {
        ops.push_back(-leftans[i]);
    }
    for (int i = 0; i < rightans.size(); i++) {
        ops.push_back(-rightans[i]);
    }
}

char s[500]; int n;

int main()
{
    ans.push_back(make_pair(0, 0));
    scanf("%d", &n);
    scanf("%s", s);
    for (int i = 0; i < (1 << n); i++) {
        for (int j = 0; j < (1 << n); j++) {
            if ((i & j) == i && s[i] == '0' && s[j] == '1') {
                printf("Impossible"); return 0;
            }
        }
    }
    for (int i = 0; i < (1 << n); i++) {
        if (s[i] == '1') continue;
        bool good = true;
        for (int j = 0; j < n; j++) {
            if ((i & (1 << j)) && s[i ^ (1 << j)] == '0') good = false;
        }
        if (!good) continue;
        vector<int> elems;
        for (int j = 0; j < n; j++) {
            if (i & (1 << j)) elems.push_back(j + 1);
        }
        vector<int> ops;
        work(elems, ops);
        for (int j = 0; j < ops.size(); j++) {
            if (ops[j] > 0) {
                cw(ops[j]);
            } else {
                ccw(-ops[j]);
            }
        }
    }
    printf("Possible\n%d\n", (int)(ans.size()) - 1);
    for (int i = 0; i < ans.size(); i++) {
        printf("%d %d\n", ans[i].first, ans[i].second);
    }
    return 0;
}

### Prompt

In cpp, your task is to solve 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 <cstdio>
#include <vector>
#include <utility>

using namespace std;

vector<pair<int, int> > ans;

void cw(int loc)
{
    for (int i = 1; i <= loc - 1; i++) {
        ans.push_back(make_pair(i, 0));
    }
    ans.push_back(make_pair(loc-1, 1));
    ans.push_back(make_pair(loc, 1));
    for (int i = loc; i >= 0; i--) {
        ans.push_back(make_pair(i, 0));
    }
}

void ccw(int loc)
{
    for (int i = 1; i <= loc; i++) {
        ans.push_back(make_pair(i, 0));
    }
    ans.push_back(make_pair(loc, 1));
    ans.push_back(make_pair(loc-1, 1));
    for (int i = loc - 1; i >= 0; i--) {
        ans.push_back(make_pair(i, 0));
    }
}

void work(vector<int>& elems, vector<int>& ops)
{
    if (elems.size() == 1) {
        ops.push_back(elems[0]); return;
    }
    vector<int> left;
    vector<int> right;
    for (int i = 0; i < elems.size() / 2; i++) {
        left.push_back(elems[i]);
    }
    for (int i = elems.size() / 2; i < elems.size(); i++) {
        right.push_back(elems[i]);
    }
    vector<int> leftans;
    vector<int> rightans;
    work(left, leftans);
    work(right, rightans);
    for (int i = 0; i < leftans.size(); i++) {
        ops.push_back(leftans[i]);
    }
    for (int i = 0; i < rightans.size(); i++) {
        ops.push_back(rightans[i]);
    }
    reverse(leftans.begin(), leftans.end());
    reverse(rightans.begin(), rightans.end());
    for (int i = 0; i < leftans.size(); i++) {
        ops.push_back(-leftans[i]);
    }
    for (int i = 0; i < rightans.size(); i++) {
        ops.push_back(-rightans[i]);
    }
}

char s[500]; int n;

int main()
{
    ans.push_back(make_pair(0, 0));
    scanf("%d", &n);
    scanf("%s", s);
    for (int i = 0; i < (1 << n); i++) {
        for (int j = 0; j < (1 << n); j++) {
            if ((i & j) == i && s[i] == '0' && s[j] == '1') {
                printf("Impossible"); return 0;
            }
        }
    }
    for (int i = 0; i < (1 << n); i++) {
        if (s[i] == '1') continue;
        bool good = true;
        for (int j = 0; j < n; j++) {
            if ((i & (1 << j)) && s[i ^ (1 << j)] == '0') good = false;
        }
        if (!good) continue;
        vector<int> elems;
        for (int j = 0; j < n; j++) {
            if (i & (1 << j)) elems.push_back(j + 1);
        }
        vector<int> ops;
        work(elems, ops);
        for (int j = 0; j < ops.size(); j++) {
            if (ops[j] > 0) {
                cw(ops[j]);
            } else {
                ccw(-ops[j]);
            }
        }
    }
    printf("Possible\n%d\n", (int)(ans.size()) - 1);
    for (int i = 0; i < ans.size(); i++) {
        printf("%d %d\n", ans[i].first, ans[i].second);
    }
    return 0;
}

```